1. Preface
2. Acknowledgements
3. Addition and Subtraction of Whole Numbers
1. Objectives
. Whole Numbers
. Reading and Writing Whole Numbers
. Rounding Whole Numbers
. Addition of Whole Numbers
. Subtraction of Whole Numbers
. Properties of Addition
. Summary of Key Concepts
. Exercise Supplement
10. Proficiency Exam
4. Multiplication and Division of Whole Numbers
1. Objectives
2. Multiplication of Whole Numbers
3. Concepts of Division of Whole Numbers
4. Division of Whole Numbers
5. Some Interesting Facts about Division
6
7.
8
WON DU BW WN
. Properties of Multiplication
. Summary of Key Concepts
. Exercise Supplement
9. Proficiency Exam
5. Exponents, Roots, and Factorization of Whole Numbers
1. Objectives
. Exponents and Roots
. Prime Factorization of Natural Numbers
. The Greatest Common Factor
. The Least Common Multiple
. Summary of Key Concepts
NOD OB WN
8. Exercise Supplement
9. Proficiency Exam
6. Introduction to Fractions and Multiplication and Division of
Fractions
1. Objectives
. Fractions of Whole Numbers
. Equivalent Fractions, Reducing Fractions to Lowest
Terms, and Raising Fractions to Higher Terms
5. Multiplication of Fractions
6. Division of Fractions
7. Applications Involving Fractions
8. Summary of Key Concepts
9. Exercise Supplement
10. Proficiency Exam
7. Addition and Subtraction of Fractions, Comparing Fractions,
and Complex Fractions
1. Objectives
2. Addition and Subtraction of Fractions with Like
Denominators
3. Addition and Subtraction of Fractions with Unlike
Denominators
. Addition and Subtraction of Mixed Numbers
. Comparing Fractions
. Complex Fractions
. Combinations of Operations with Fractions
. Summary of Key Concepts
. Exercise Supplement
10. Proficiency Exam
8. Decimals
1. Objectives
2. Reading and Writing Decimals
WON MD Uf
. Converting a Decimal to a Fraction
. Rounding Decimals
. Addition and Subtraction of Decimals
. Multiplication of Decimals
. Division of Decimals
. Nonterminating Divisions
. Converting a Fraction to a Decimal
. Combinations of Operations with Decimals and Fractions
. Summary of Key Concepts
. Exercise Supplement
13.
Proficiency Exam
9. Ratios and Rates
. Objectives
. Ratios and Rates
. Proportions
. Applications of Proportions
. Percent
. Fractions of One Percent
. Applications of Percents
. Summary of Key Concepts
. Exercise Supplement
10.
Proficiency Exam
10. Techniques of Estimation
1.
. Estimation by Rounding
. Estimation by Clustering
. Mental Arithmetic-Using the Distributive Property
. Estimation by Rounding Fractions
. Summary of Key Concepts
. Exercise Supplement
8.
NOD OB WN
Objectives
Proficiency Exam
11. Measurement and Geometry
CON DU BWN FE
. Objectives
. Measurement and the United States System
. The Metric System of Measurement
. Simplification of Denominate Numbers
. Perimeter and Circumference of Geometric Figures
. Area and Volume of Geometric Figures and Objects
. Summary of Key Concepts
. Exercise Supplement
9.
Proficiency Exam
12. Signed Numbers
.,
. Variables, Constants, and Real Numbers
. Signed Numbers
. Absolute Value
. Addition of Signed Numbers
. Subtraction of Signed Numbers
. Multiplication and Division of Signed Numbers
. Summary of Key Concepts
. Exercise Supplement
10.
WON ADU BW WN
Objectives
Proficiency Exam
13. Algebraic Expressions and Equations
. Objectives
. Algebraic Expressions
. Combining Like Terms Using Addition and Subtraction
. Solving Equations of the Form x+a=b and x-a=b
. Solving Equations of the Form ax=b and x/a=b
. Applications I: Translating Words to Mathematical
Symbols
. Applications Il: Solving Problems
. Summary of Key Concepts
. Exercise Supplement
. Proficiency Exam
Preface
This module contains the preface for Fundamentals of Mathematics by
Denny Burzynski and Wade Ellis, Jr.
To the next generation of explorers: Kristi, BreAnne, Lindsey, Randi, Piper,
Meghan, Wyatt, Lara, Mason, and Sheanna.
Fundamentals of Mathematics is a work text that covers the traditional
topics studied in a modern prealgebra course, as well as the topics of
estimation, elementary analytic geometry, and introductory algebra. It is
intended for students who
1. have had a previous course in prealgebra,
2. wish to meet the prerequisite of a higher level course such as
elementary algebra, and
3. need to review fundamental mathematical concepts and techniques.
This text will help the student develop the insight and intuition necessary to
master arithmetic techniques and manipulative skills. It was written with the
following main objectives:
1. to provide the student with an understandable and usable source of
information,
2. to provide the student with the maximum opportunity to see that
arithmetic concepts and techniques are logically based,
3. to instill in the student the understanding and intuitive skills necessary
to know how and when to use particular arithmetic concepts in
subsequent material, courses, and nonclassroom situations, and
4. to give the student the ability to correctly interpret arithmetically
obtained results.
We have tried to meet these objectives by presenting material dynamically,
much the way an instructor might present the material visually in a
classroom. (See the development of the concept of addition and subtraction
of fractions in [link], for example.) Intuition and understanding are some of
the keys to creative thinking; we believe that the material presented in this
text will help the student realize that mathematics is a creative subject.
This text can be used in standard lecture or self-paced classes. To help meet
our objectives and to make the study of prealgebra a pleasant and rewarding
experience, Fundamentals of Mathematics is organized as follows.
Pedagogical Features
The work text format gives the student space to practice mathematical skills
with ready reference to sample problems. The chapters are divided into
sections, and each section is a complete treatment of a particular topic,
which includes the following features:
e Section Overview
¢ Sample Sets
e Practice Sets
e Section Exercises
e Exercises for Review
e Answers to Practice Sets
The chapters begin with Objectives and end with a Summary _of Key.
Concepts, an Exercise Supplement, and a Proficiency Exam.
Objectives
Each chapter begins with a set of objectives identifying the material to be
covered. Each section begins with an overview that repeats the objectives
for that particular section. Sections are divided into subsections that
correspond to the section objectives, which makes for easier reading.
Sample Sets
Fundamentals of Mathematics contains examples that are set off in boxes
for easy reference. The examples are referred to as Sample Sets for two
reasons:
1. They serve as a representation to be imitated, which we believe will
foster understanding of mathematical concepts and provide experience
with mathematical techniques.
2. Sample Sets also serve as a preliminary representation of problem-
solving techniques that may be used to solve more general and more
complicated problems.
The examples have been carefully chosen to illustrate and develop concepts
and techniques in the most instructive, easily remembered way. Concepts
and techniques preceding the examples are introduced at a level below that
normally used in similar texts and are thoroughly explained, assuming little
previous knowledge.
Practice Sets
A parallel Practice Set follows each Sample Set, which reinforces the
concepts just learned. There is adequate space for the student to work each
problem directly on the page.
Answers to Practice Sets
The Answers to Practice Sets are given at the end of each section and can
be easily located by referring to the page number, which appears after the
last Practice Set in each section.
Section Exercises
The exercises at the end of each section are graded in terms of difficulty,
although they are not grouped into categories. There is an ample number of
problems, and after working through the exercises, the student will be
capable of solving a variety of challenging problems.
The problems are paired so that the odd-numbered problems are equivalent
in kind and difficulty to the even-numbered problems. Answers to the odd-
numbered problems are provided at the back of the book.
Exercises for Review
This section consists of five problems that form a cumulative review of the
material covered in the preceding sections of the text and is not limited to
material in that chapter. The exercises are keyed by section for easy
reference. Since these exercises are intended for review only, no work space
is provided.
Summary of Key Concepts
A summary of the important ideas and formulas used throughout the
chapter is included at the end of each chapter. More than just a list of terms,
the summary is a valuable tool that reinforces concepts in preparation for
the Proficiency Exam at the end of the chapter, as well as future exams. The
summary keys each item to the section of the text where it is discussed.
Exercise Supplement
In addition to numerous section exercises, each chapter includes
approximately 100 supplemental problems, which are referenced by
section. Answers to the odd-numbered problems are included in the back of
the book.
Proficiency Exam
Each chapter ends with a Proficiency Exam that can serve as a chapter
review or evaluation. The Proficiency Exam is keyed to sections, which
enables the student to refer back to the text for assistance. Answers to all
the problems are included in the Answer Section at the end of the book.
Content
The writing style used in Fundamentals of Mathematics is informal and
friendly, offering a straightforward approach to prealgebra mathematics. We
have made a deliberate effort not to write another text that minimizes the
use of words because we believe that students can best study arithmetic
concepts and understand arithmetic techniques by using words and symbols
rather than symbols alone. It has been our experience that students at the
prealgebra level are not nearly experienced enough with mathematics to
understand symbolic explanations alone; they need literal explanations to
guide them through the symbols.
We have taken great care to present concepts and techniques so they are
understandable and easily remembered. After concepts have been
developed, students are warned about common pitfalls. We have tried to
make the text an information source accessible to prealgebra students.
Addition and Subtraction of Whole Numbers
This chapter includes the study of whole numbers, including a discussion of
the Hindu-Arabic numeration and the base ten number systems. Rounding
whole numbers is also presented, as are the commutative and associative
properties of addition.
Multiplication and Division of Whole Numbers
The operations of multiplication and division of whole numbers are
explained in this chapter. Multiplication is described as repeated addition.
Viewing multiplication in this way may provide students with a
visualization of the meaning of algebraic terms such as 8a when they start
learning algebra. The chapter also includes the commutative and associative
properties of multiplication.
Exponents, Roots, and Factorizations of Whole Numbers
The concept and meaning of the word root is introduced in this chapter. A
method of reading root notation and a method of determining some
common roots, both mentally and by calculator, is then presented. We also
present grouping symbols and the order of operations, prime factorization
of whole numbers, and the greatest common factor and least common
multiple of a collection of whole numbers.
Introduction to Fractions and Multiplication and Division of Fractions
We recognize that fractions constitute one of the foundations of problem
solving. We have, therefore, given a detailed treatment of the operations of
multiplication and division of fractions and the logic behind these
operations. We believe that the logical treatment and many practice
exercises will help students retain the information presented in this chapter
and enable them to use it as a foundation for the study of rational
expressions in an algebra course.
Addition and Subtraction of Fractions, Comparing Fractions, and
Complex Fractions
A detailed treatment of the operations of addition and subtraction of
fractions and the logic behind these operations is given in this chapter.
Again, we believe that the logical treatment and many practice exercises
will help students retain the information, thus enabling them to use it in the
study of rational expressions in an algebra course. We have tried to make
explanations dynamic. A method for comparing fractions is introduced,
which gives the student another way of understanding the relationship
between the words denominator and denomination. This method serves to
show the student that it is sometimes possible to compare two different
types of quantities. We also study a method of simplifying complex
fractions and of combining operations with fractions.
Decimals
The student is introduced to decimals in terms of the base ten number
system, fractions, and digits occurring to the right of the units position. A
method of converting a fraction to a decimal is discussed. The logic behind
the standard methods of operating on decimals is presented and many
examples of how to apply the methods are given. The word of as related to
the operation of multiplication is discussed. Nonterminating divisions are
examined, as are combinations of operations with decimals and fractions.
Ratios and Rates
We begin by defining and distinguishing the terms ratio and rate. The
meaning of proportion and some applications of proportion problems are
described. Proportion problems are solved using the "Five-Step Method."
We hope that by using this method the student will discover the value of
introducing a variable as a first step in problem solving and the power of
organization. The chapter concludes with discussions of percent, fractions
of one percent, and some applications of percent.
Techniques of Estimation
One of the most powerful problem-solving tools is a knowledge of
estimation techniques. We feel that estimation is so important that we
devote an entire chapter to its study. We examine three estimation
techniques: estimation by rounding, estimation by clustering, and
estimation by rounding fractions. We also include a section on the
distributive property, an important algebraic property.
Measurement and Geometry
This chapter presents some of the techniques of measurement in both the
United States system and the metric system. Conversion from one unit to
another (in a system) is examined in terms of unit fractions. A discussion of
the simplification of denominate numbers is also included. This discussion
helps the student understand more clearly the association between pure
numbers and dimensions. The chapter concludes with a study of perimeter
and circumference of geometric figures and area and volume of geometric
figures and objects.
Signed Numbers
A look at algebraic concepts and techniques is begun in this chapter. Basic
to the study of algebra is a working knowledge of signed numbers.
Definitions of variables, constants, and real numbers are introduced. We
then distinguish between positive and negative numbers, learn how to read
signed numbers, and examine the origin and use of the double-negative
property of real numbers. The concept of absolute value is presented both
geometrically (using the number line) and algebraically. The algebraic
definition is followed by an interpretation of its meaning and several
detailed examples of its use. Addition, subtraction, multiplication, and
division of signed numbers are presented first using the number line, then
with absolute value.
Algebraic Expressions and Equations
The student is introduced to some elementary algebraic concepts and
techniques in this final chapter. Algebraic expressions and the process of
combining like terms are discussed in [link] and [link]. The method of
combining like terms in an algebraic expression is explained by using the
interpretation of multiplication as a description of repeated addition (as in
[link]).
Acknowledgements
This module contains the authors' acknowledgments and dedication of the
book, Fundamentals of Mathematics by Denny Burzynski and Wade Ellis.
Many extraordinarily talented people are responsible for helping to create
this text. We wish to acknowledge the efforts and skill of the following
mathematicians. Their contributions have been invaluable.
e Barbara Conway, Berkshire Community College
e Bill Hajdukiewicz, Miami-Dade Community College
e Virginia Hamilton, Shawnee State University
e David Hares, El Centro College
e Norman Lee, Ball State University
e Ginger Y. Manchester, Hinds Junior College
e John R. Martin, Tarrant County Junior College
e Shelba Mormon, Northlake College
e Lou Ann Pate, Pima Community College
e Gus Pekara, Oklahoma City Community College
e David Price, Tarrant County Junior College
¢ David Schultz, Virginia Western Community College
e Sue S. Watkins, Lorain County Community College
e Elizabeth M. Wayt, Tennessee State University
e Prentice E. Whitlock, Jersey City State College
¢ Thomas E. Williamson, Montclair State College
Special thanks to the following individuals for their careful accuracy
reviews of manuscript, galleys, and page proofs: Steve Blasberg, West
Valley College; Wade Ellis, Sr., University of Michigan; John R. Martin,
Tarrant County Junior College; and Jane Ellis. We would also like to thank
Amy Miller and Guy Sanders, Branham High School.
Our sincere thanks to Debbie Wiedemann for her encouragement,
suggestions concerning psychobiological examples, proofreading much of
the manuscript, and typing many of the section exercises; Sandi Wiedemann
for collating the annotated reviews, counting the examples and exercises,
and untiring use of "white-out"; and Jane Ellis for solving and typing all of
the exercise solutions.
We thank the following people for their excellent work on the various
ancillary items that accompany Fundamentals of Mathematics: Steve
Blasberg, West Valley College; Wade Ellis, Sr., University of Michigan; and
Jane Ellis ( Instructor's Manual); John R. Martin, Tarrant County Junior
College (Student Solutions Manual and Study Guide); Virginia Hamilton,
Shawnee State University (Computerized Test Bank); Patricia Morgan, San
Diego State University (Prepared Tests); and George W. Bergeman,
Northern Virginia Community College (Maxis Interactive Software).
We also thank the talented people at Saunders College Publishing whose
efforts made this text run smoothly and less painfully than we had
imagined. Our particular thanks to Bob Stern, Mathematics Editor, Ellen
Newman, Developmental Editor, and Janet Nuciforo, Project Editor. Their
guidance, suggestions, open minds to our suggestions and concerns, and
encouragement have been extraordinarily helpful. Although there were
times we thought we might be permanently damaged from rereading and
rewriting, their efforts have improved this text immensely. It is a pleasure to
work with such high-quality professionals.
Denny Burzynski
Wade Ellis, Jr.
San Jose, California
December 1988
I would like to thank Doug Campbell, Ed Lodi, and Guy Sanders for
listening to my frustrations and encouraging me on. Thanks also go to my
cousin, David Raffety, who long ago in Sequoia National Forest told me
what a differential equation is.
Particular thanks go to each of my colleagues at West Valley College. Our
everyday conversations regarding mathematics instruction have been of the
utmost importance to the development of this text and to my teaching
career.
D.B.
Objectives
This module contains the learning objectives for the chapter "Addition and
Subtraction of Whole Numbers" from Fundamentals of Mathematics by
Denny Burzynski and Wade Ellis, jr.
After completing this chapter, you should
Whole Numbers ({link])
e know the difference between numbers and numerals
e know why our number system is called the Hindu-Arabic numeration
system
e understand the base ten positional number system
e be able to identify and graph whole numbers
Reading and Writing Whole Numbers ([{link])
e be able to read and write a whole number
Rounding Whole Numbers ((link])
¢ understand that rounding is a method of approximation
e be able to round a whole number to a specified position
Addition of Whole Numbers ({link])
e understand the addition process
e be able to add whole numbers
e be able to use the calculator to add one whole number to another
Subtraction of Whole Numbers ({link])
e understand the subtraction process
e be able to subtract whole numbers
e be able to use a calculator to subtract one whole number from another
whole number
Properties of Addition ({link])
e understand the commutative and associative properties of addition
¢ understand why 0 is the additive identity
Whole Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses many of aspects of whole
numbers, including the Hindu-Arabic numeration system, the base ten
positional number system, and the graphing of whole numbers. By the end
of this module students should be able to: know the difference between
numbers and numerals, know why our number system is called the Hindu-
Arabic numeration system, understand the base ten positional number
system, and identify and graph whole numbers.
Section Overview
e Numbers and Numerals
e The Hindu-Arabic Numeration System
e The Base Ten Positional Number System
¢ Whole Numbers
¢ Graphing Whole Numbers
Numbers and Numerals
We begin our study of introductory mathematics by examining its most
basic building block, the number.
Number
A number is a concept. It exists only in the mind.
The earliest concept of a number was a thought that allowed people to
mentally picture the size of some collection of objects. To write down the
number being conceptualized, a numeral is used.
Numeral
A numeral is a symbol that represents a number.
In common usage today we do not distinguish between a number and a
numeral. In our study of introductory mathematics, we will follow this
common usage.
Sample Set A
The following are numerals. In each case, the first represents the number
four, the second represents the number one hundred twenty-three, and the
third, the number one thousand five. These numbers are represented in
different ways.
e Hindu-Arabic numerals
4, 123, 1005
e Roman numerals
IV, CX XIII, MV
e Egyptian numerals
li dd> ONNI11> biti
Strokes Coiled rope, Lotus flower
heel bones, and strokes
and strokes
Practice Set A
Exercise:
Problem:
Wo
Do the phrases "four," "one hundred twenty-three," and "one thousand
five" qualify as numerals? Yes or no?
Solution:
Yes. Letters are symbols. Taken as a collection (a written word), they
represent a number.
The Hindu-Arabic Numeration System
Hindu-Arabic Numeration System
Our society uses the Hindu-Arabic numeration system. This system of
numeration began shortly before the third century when the Hindus
invented the numerals
0123456789
Leonardo Fibonacci
About a thousand years later, in the thirteenth century, a mathematician
named Leonardo Fibonacci of Pisa introduced the system into Europe. It
was then popularized by the Arabs. Thus, the name, Hindu-Arabic
numeration system.
The Base Ten Positional Number System
Digits
The Hindu-Arabic numerals 01234567 89 are called digits. We can
form any number in the number system by selecting one or more digits and
placing them in certain positions. Each position has a particular value. The
Hindu mathematician who devised the system about A.D. 500 stated that
"from place to place each is ten times the preceding."
Base Ten Positional Systems
It is for this reason that our number system is called a positional number
system with base ten.
Commas
When numbers are composed of more than three digits, commas are
sometimes used to separate the digits into groups of three.
Periods
These groups of three are called periods and they greatly simplify reading
numbers.
In the Hindu-Arabic numeration system, a period has a value assigned to
each or its three positions, and the values are the same for each period. The
position values are
Thus, each period contains a position for the values of one, ten, and
hundred. Notice that, in looking from right to left, the value of each position
is ten times the preceding. Each period has a particular name.
ae | eee | ee ee) ee | ee ee eee
ee Oe |
Trillions Billions Millions Thousands Units
As we continue from right to left, there are more periods. The five periods
listed above are the most common, and in our study of introductory
mathematics, they are sufficient.
The following diagram illustrates our positional number system to trillions.
(There are, to be sure, other periods.)
In our positional number system, the value of a digit is determined by its
position in the number.
Sample Set B
Example:
Find the value of 6 in the number 7,261.
Since 6 is in the tens position of the units period, its value is 6 tens.
6 tens = 60
Example:
Find the value of 9 in the number 86,932,106,005.
Since 9 is in the hundreds position of the millions period, its value is 9
hundred millions.
9 hundred millions = 9 hundred million
Example:
Find the value of 2 in the number 102,001.
Since 2 is in the ones position of the thousands period, its value is 2 one
thousands.
2 one thousands = 2 thousand
Practice Set B
Exercise:
Problem: Find the value of 5 in the number 65,000.
Solution:
five thousand
Exercise:
Problem: Find the value of 4 in the number 439,997,007,010.
Solution:
four hundred billion
Exercise:
Problem: Find the value of 0 in the number 108.
Solution:
zero tens, Or Zero
Whole Numbers
Whole Numbers
Numbers that are formed using only the digits
0123456789
are called whole numbers. They are
On 4, 25354) 5,G,-750 79, 10.11. 120138. 14. Toya
The three dots at the end mean "and so on in this same pattern."
Graphing Whole Numbers
Number Line
Whole numbers may be visualized by constructing a number line. To
construct a number line, we simply draw a straight line and choose any
point on the line and label it 0.
Origin
This point is called the origin. We then choose some convenient length, and
moving to the right, mark off consecutive intervals (parts) along the line
Starting at 0. We label each new interval endpoint with the next whole
number.
Graphing
We can visually display a whole number by drawing a closed circle at the
point labeled with that whole number. Another phrase for visually
displaying a whole number is graphing the whole number. The word graph
means to "visually display."
Sample Set C
Example:
Graph the following whole numbers: 3, 5, 9.
Example:
Specify the whole numbers that are graphed on the following number line.
The break in the number line indicates that we are aware of the whole
numbers between 0 and 106, and 107 and 872, but we are not listing them
due to space limitations.
0 106 107 872 873 874
The numbers that have been graphed are
0, 106, 873, 874
Practice Set C
Exercise:
Problem: Graph the following whole numbers: 46, 47, 48, 325, 327.
“2 “"/ ps —-
Solution:
0 46 47 48 325 326 327
Exercise:
Problem:
Specify the whole numbers that are graphed on the following number
line.
0 123 4 5 6 112 113 978 979
Solution:
4,5, 6, 113, 978
A line is composed of an endless number of points. Notice that we have
labeled only some of them. As we proceed, we will discover new types of
numbers and determine their location on the number line.
Exercises
Exercise:
Problem: What is a number?
Solution:
concept
Exercise:
Problem: What is a numeral?
Exercise:
Problem: Does the word "eleven" qualify as a numeral?
Solution:
Yes, since it is a symbol that represents a number.
Exercise:
Problem: How many different digits are there?
Exercise:
Problem:
Our number system, the Hindu-Arabic number system, is a number
system with base .
Solution:
positional; 10
Exercise:
Problem:
Numbers composed of more than three digits are sometimes separated
into groups of three by commas. These groups of three are called .
Exercise:
Problem:
In our number system, each period has three values assigned to it.
These values are the same for each period. From right to left, what are
they?
Solution:
ones, tens, hundreds
Exercise:
Problem:
Each period has its own particular name. From right to left, what are
the names of the first four?
Exercise:
Problem: In the number 841, how many tens are there?
Solution:
4
Exercise:
Problem: In the number 3,392, how many ones are there?
Exercise:
Problem: In the number 10,046, how many thousands are there?
Solution:
0
Exercise:
Problem:
In the number 779,844,205, how many ten millions are there?
Exercise:
Problem:
In the number 65,021, how many hundred thousands are there?
Solution:
0
For following problems, give the value of the indicated digit in the given
number.
Exercise:
Problem: 5 in 599
Exercise:
Problem: 1 in 310,406
Solution:
ten thousand
Exercise:
Problem: 9 in 29,827
Exercise:
Problem: 6 in 52,561,001,100
Solution:
6 ten millions = 60 million
Exercise:
Problem
Write a two-digit number that has an eight in the tens position.
Exercise:
Problem
Write a four-digit number that has a one in the thousands position and
a zero in
the ones position.
Solution:
1,340 (answers may vary)
Exercise:
Problem
Exercise:
Problem
: How many two-digit whole numbers are there?
: How many three-digit whole numbers are there?
Solution:
900
Exercise:
Problem
Exercise:
Problem
: How many four-digit whole numbers are there?
: Is there a smallest whole number? If so, what is it?
Solution:
yes; Zero
Exercise:
Problem: Is there a largest whole number? If so, what is it?
Exercise:
Problem: Another term for "visually displaying" is .
Solution:
graphing
Exercise:
Problem: The whole numbers can be visually displayed ona.
Exercise:
Problem:
Graph (visually display) the following whole numbers on the number
line below: 0, 1, 31, 34.
0 12 8 4 #‘\%Y 29 30 31 32 33 34
Solution:
0 1 2 83 4 2 30 31 32 33 34
Exercise:
Problem:
Construct a number line in the space provided below and graph
(visually display) the following whole numbers: 84, 85, 901, 1006,
1007.
Exercise:
Problem:
Specify, if any, the whole numbers that are graphed on the following
number line.
tJ tt MN ot
0 61 62 63 64 99 100 101 102
Solution:
61, 99, 100, 102
Exercise:
Problem:
Specify, if any, the whole numbers that are graphed on the following
number line.
+ —+-A\— ++ + + “| -— + 4+ A +
01 8 9 10 11 73 74 85 86 87
Reading and Writing Whole Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to read and write whole
numbers. By the end of this module, students should be able to read and
write whole numbers.
Section Overview
e Reading Whole Numbers
e Writing Whole Numbers
Because our number system is a positional number system, reading and
writing whole numbers is quite simple.
Reading Whole Numbers
To convert a number that is formed by digits into a verbal phrase, use the
following method:
1. Beginning at the right and working right to left, separate the number
into distinct periods by inserting commas every three digits.
2. Beginning at the left, read each period individually, saying the period
name.
Sample Set A
Write the following numbers as words.
Example:
Read 42958.
1. Beginning at the right, we can separate this number into distinct
periods by inserting a comma between the 2 and 9.
42,958
2. Beginning at the left, we read each period individually:
Lowa alt es , —— Forty-two thousand
Le cceaclnciceninsimeniianl
Thousands period
[a 5) 8, —— nine hundred fifty-eight
Forty-two thousand, nine hundred fifty-eight.
Example:
Read 307991343.
1. Beginning at the right, we can separate this number into distinct
periods by placing commas between the 1 and 3 and the 7 and 9.
307,991,343
2. Beginning at the left, we read each period individually.
| ‘aes
L_jLojJt 1, ——» Three hundred seven million,
eel
Millions period
bat ear hat » —— nine hundred ninety-one thousand,
Li eenacnemainadianl
Thousands period
3, el 3, —— three hundred forty-three
Lf
Units period
Three hundred seven million, nine hundred ninety-one thousand, three
hundred forty-three.
Example:
Read 36000000000001.
1. Beginning at the right, we can separate this number into distinct
periods by placing commas. 36,000,000,001
2. Beginning at the left, we read each period individually.
a | 3 8, , —— Thirty-six trillion,
LS
Trillions period
0D: @ 6 aon
L_jLUJL_J,—— zero billion,
Le
Billions period
0,0,,0 _
LJLLjLOJ,—- zero million,
LS
Millions period
0 ,0,,0
L-JL JL J,-—— zero thousand,
—————: |
Thousands period
"al el | ek
LE
Units period
Thirty-six trillion, one.
Practice Set A
Write each number in words.
Exercise:
Problem: 12,542
Solution:
Twelve thousand, five hundred forty-two
Exercise:
Problem: 101,074,003
Solution:
One hundred one million, seventy-four thousand, three
Exercise:
Problem: 1,000,008
Solution:
One million, eight
Writing Whole Numbers
To express a number in digits that is expressed in words, use the following
method:
1. Notice first that a number expressed as a verbal phrase will have its
periods set off by commas.
2. Starting at the beginning of the phrase, write each period of numbers
individually.
3. Using commas to separate periods, combine the periods to form one
number.
Sample Set B
Write each number using digits.
Example:
Seven thousand, ninety-two.
Using the comma as a period separator, we have
Seven thousand , —— 7,
nearereete eae
ninety-two, ——> 092
7,092
Example:
Fifty billion, one million, two hundred thousand, fourteen.
Using the commas as period separators, we have
Fifty billion,, —— 50,
one million ,—— 001,
two hundred thousand , —— 200,
fourteen, —— 014
50,001,200,014
Example:
Ten million, five hundred twelve.
The comma sets off the periods. We notice that there is no thousands
period. We'll have to insert this ourselves.
Ten million ,—— 10,
zero thousand , —— 000,
five hundred twelve, —— 512
ee
10,000,512
Practice Set B
Express each number using digits.
Exercise:
Problem: One hundred three thousand, twenty-five.
Solution:
103,025
Exercise:
Problem: Six million, forty thousand, seven.
Solution:
6,040,007
Exercise:
Problem:
Twenty trillion, three billion, eighty million, one hundred nine
thousand, four hundred two.
Solution:
20,003,080, 109,402
Exercise:
Problem: Eighty billion, thirty-five.
Solution:
80,000,000,035
Exercises
For the following problems, write all numbers in words.
Exercise:
Problem: 912
Solution:
nine hundred twelve
Exercise:
Problem: 84
Exercise:
Problem: 1491
Solution:
one thousand, four hundred ninety-one
Exercise:
Problem: 8601
Exercise:
Problem: 35,223
Solution:
thirty-five thousand, two hundred twenty-three
Exercise:
Problem: 71,006
Exercise:
Problem: 437,105
Solution:
four hundred thirty-seven thousand, one hundred five
Exercise:
Problem: 201,040
Exercise:
Problem: 8,001,001
Solution:
eight million, one thousand, one
Exercise:
Problem: 16,000,053
Exercise:
Problem: 770,311,101
Solution:
seven hundred seventy million, three hundred eleven thousand, one
hundred one
Exercise:
Problem: 83,000,000,007
Exercise:
Problem: 106,100,001,010
Solution:
one hundred six billion, one hundred million, one thousand ten
Exercise:
Problem: 3,333,444,777
Exercise:
Problem: 800,000,800,000
Solution:
eight hundred billion, eight hundred thousand
Exercise:
Problem:
A particular community college has 12,471 students enrolled.
Exercise:
Problem:
A person who watches 4 hours of television a day spends 1460 hours a
year watching T.V.
Solution:
four; one thousand, four hundred sixty
Exercise:
Problem:
Astronomers believe that the age of the earth is about 4,500,000,000
years.
Exercise:
Problem:
Astronomers believe that the age of the universe is about
20,000,000,000 years.
Solution:
twenty billion
Exercise:
Problem:
There are 9690 ways to choose four objects from a collection of 20.
Exercise:
Problem:
If a 412 page book has about 52 sentences per page, it will contain
about 21,424 sentences.
Solution:
four hundred twelve; fifty-two; twenty-one thousand, four hundred
twenty-four
Exercise:
Problem:
In 1980, in the United States, there was $1,761,000,000,000 invested
in life insurance.
Exercise:
Problem:
In 1979, there were 85,000 telephones in Alaska and 2,905,000
telephones in Indiana.
Solution:
one thousand, nine hundred seventy-nine; eighty-five thousand; two
million, nine hundred five thousand
Exercise:
Problem:
In 1975, in the United States, it is estimated that 52,294,000 people
drove to work alone.
Exercise:
Problem:
In 1980, there were 217 prisoners under death sentence that were
divorced.
Solution:
one thousand, nine hundred eighty; two hundred seventeen
Exercise:
Problem:
In 1979, the amount of money spent in the United States for regular-
session college education was $50,721,000,000,000.
Exercise:
Problem:
In 1981, there were 1,956,000 students majoring in business in U.S.
colleges.
Solution:
one thousand, nine hundred eighty one; one million, nine hundred
fifty-six thousand
Exercise:
Problem:
In 1980, the average fee for initial and follow up visits to a medical
doctors office was about $34.
Exercise:
Problem:
In 1980, there were approximately 13,100 smugglers of aliens
apprehended by the Immigration border patrol.
Solution:
one thousand, nine hundred eighty; thirteen thousand, one hundred
Exercise:
Problem:
In 1980, the state of West Virginia pumped 2,000,000 barrels of crude
oil, whereas Texas pumped 975,000,000 barrels.
Exercise:
Problem: The 1981 population of Uganda was 12,630,000 people.
Solution:
twelve million, six hundred thirty thousand
Exercise:
Problem:
In 1981, the average monthly salary offered to a person with a Master's
degree in mathematics was $1,685.
For the following problems, write each number using digits.
Exercise:
Problem: Six hundred eighty-one
Solution:
681
Exercise:
Problem: Four hundred ninety
Exercise:
Problem: Seven thousand, two hundred one
Solution:
F201
Exercise:
Problem: Nineteen thousand, sixty-five
Exercise:
Problem: Five hundred twelve thousand, three
Solution:
512,003
Exercise:
Problem:
Two million, one hundred thirty-three thousand, eight hundred fifty-
nine
Exercise:
Problem: Thirty-five million, seven thousand, one hundred one
Solution:
35,007,101
Exercise:
Problem: One hundred million, one thousand
Exercise:
Problem: Sixteen billion, fifty-nine thousand, four
Solution:
16,000,059,004
Exercise:
Problem:
Nine hundred twenty billion, four hundred seventeen million, twenty-
one thousand
Exercise:
Problem: Twenty-three billion
Solution:
23,000,000,000
Exercise:
Problem:
Fifteen trillion, four billion, nineteen thousand, three hundred five
Exercise:
Problem: One hundred trillion, one
Solution:
100,000,000,000,001
Exercises for Review
Exercise:
Problem: ({link]) How many digits are there?
Exercise:
Problem: ({link]) In the number 6,641, how many tens are there?
Solution:
4
Exercise:
Problem: ({link]) What is the value of 7 in 44,763?
Exercise:
Problem: ({link]) Is there a smallest whole number? If so, what is it?
Solution:
yes, Zero
Exercise:
Problem:
({link]) Write a four-digit number with a 9 in the tens position.
Rounding Whole Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis,
Jr. This module discusses how to round whole numbers. By the end of the module students
should be able to understand that rounding is a method of approximation and round a
whole number to a specified position.
Section Overview
e Rounding as an Approximation
e The Method of Rounding Numbers
Rounding as an Approximation
A primary use of whole numbers is to keep count of how many objects there are in a
collection. Sometimes we're only interested in the approximate number of objects in the
collection rather than the precise number. For example, there are approximately 20
symbols in the collection below.
a Pi a x Pi Pi a a
x, eF &
The precise number of symbols in the above collection is 18.
Rounding
We often approximate the number of objects in a collection by mentally seeing the
collection as occurring in groups of tens, hundreds, thousands, etc. This process of
approximation is called rounding. Rounding is very useful in estimation. We will study
estimation in Chapter 8.
When we think of a collection as occurring in groups of tens, we say we're rounding to the
nearest ten. When we think of a collection as occurring in groups of hundreds, we say
we're rounding to the nearest hundred. This idea of rounding continues through thousands,
ten thousands, hundred thousands, millions, etc.
The process of rounding whole numbers is illustrated in the following examples.
Example:
Round 67 to the nearest ten.
On the number line, 67 is more than halfway from 60 to 70. The digit immediately to the
right of the tens digit, the round-off digit, is the indicator for this.
6 — - tens
0 60 67 38670
67 is closer to 7 tens
than it is to 6 tens.
Thus, 67, rounded to the nearest ten, is 70.
Example:
Round 4,329 to the nearest hundred.
On the number line, 4,329 is less than halfway from 4,300 to 4,400. The digit to the
immediate right of the hundreds digit, the round-off digit, is the indicator.
3 ices | ie hundreds
0 4,300 4,329 4,400
4,329 is closer to 43 hundreds
than it is to 44 hundreds.
Thus, 4,329, rounded to the nearest hundred is 4,300.
Example:
Round 16,500 to the nearest thousand.
On the number line, 16,500 is exactly halfway from 16,000 to 17,000.
6 thousands 7% is 7 thousands
0 16,000 16,500 17,000
By convention, when the number to be rounded is exactly halfway between two numbers,
it is rounded to the higher number.
Thus, 16,500, rounded to the nearest thousand, is 17,000.
Example:
A person whose salary is $41,450 per year might tell a friend that she makes $41,000 per
year. She has rounded 41,450 to the nearest thousand. The number 41,450 is closer to
41,000 than it is to 42,000.
The Method of Rounding Whole Numbers
From the observations made in the preceding examples, we can use the following method
to round a whole number to a particular position.
1. Mark the position of the round-off digit.
2. Note the digit to the immediate right of the round-off digit.
a. If it is less than 5, replace it and all the digits to its right with zeros. Leave the
round-off digit unchanged.
b. If it is 5 or larger, replace it and all the digits to its right with zeros. Increase the
round-off digit by 1.
Sample Set A
Use the method of rounding whole numbers to solve the following problems.
Example:
Round 3,426 to the nearest ten.
1. We are rounding to the tens position. Mark the digit in the tens position
3,426
f
tens position
2. Observe the digit immediately to the right of the tens position. It is 6. Since 6 is
greater than 5, we round up by replacing 6 with 0 and adding 1 to the digit in the tens
position (the round-off position): 2+ 1= 3.
3,430
Thus, 3,426 rounded to the nearest ten is 3,430.
Example:
Round 9,614,018,007 to the nearest ten million.
1. We are rounding to the nearest ten million.
9,614,018,007
t
ten millions position
2. Observe the digit immediately to the right of the ten millions position. It is 4. Since 4
is less than 5, we round down by replacing 4 and all the digits to its right with zeros.
9,610,000,000
Thus, 9,614,018,007 rounded to the nearest ten million is 9,610,000,000.
Example:
Round 148,422 to the nearest million.
1. Since we are rounding to the nearest million, we'll have to imagine a digit in the
millions position. We'll write 148,422 as 0,148,422.
0,148,422
millions position
2. The digit immediately to the right is 1. Since 1 is less than 5, we'll round down by
replacing it and all the digits to its right with zeros.
0,000,000
This number is 0.
Thus, 148,422 rounded to the nearest million is 0.
Example:
Round 397,000 to the nearest ten thousand.
1. We are rounding to the nearest ten thousand.
397,000
ten thousand position
2. The digit immediately to the right of the ten thousand position is 7. Since 7 is greater
than 5, we round up by replacing 7 and all the digits to its right with zeros and
adding 1 to the digit in the ten thousands position. But 9 + 1 = 10 and we must
carry the 1 to the next (the hundred thousands) position.
400,000
Thus, 397,000 rounded to the nearest ten thousand is 400,000.
Practice Set A
Use the method of rounding whole numbers to solve each problem.
Exercise:
Problem: Round 3387 to the nearest hundred.
Solution:
3400
Exercise:
Problem: Round 26,515 to the nearest thousand.
Solution:
27,000
Exercise:
Problem: Round 30,852,900 to the nearest million.
Solution:
31,000,000
Exercise:
Problem: Round 39 to the nearest hundred.
Solution:
0
Exercise:
Problem: Round 59,600 to the nearest thousand.
Solution:
60,000
Exercises
For the following problems, complete the table by rounding each number to the indicated
positions.
Exercise:
Problem: 1,642
hundred thousand ten thousand million
Solution:
hundred thousand ten thousand million
1,600 2000 0 0
Exercise:
Problem: 5,221
hundred thousand
Exercise:
Problem: 91,803
Hundred thousand
Solution:
Hundred thousand
91,800 92,000
Exercise:
Problem: 106,007
hundred thousand
Exercise:
ten thousand
ten thousand
ten thousand
90,000
ten thousand
million
million
million
0
million
Problem: 208
hundred thousand ten thousand
Solution:
hundred thousand ten thousand
200 0 0
Exercise:
Problem: 199
hundred thousand ten thousand
Exercise:
Problem: 863
million
million
0
million
hundred thousand
Solution:
hundred thousand
900 1,000
Exercise:
Problem: 794
hundred thousand
Exercise:
Problem: 925
hundred thousand
Solution:
ten thousand
ten thousand
0
ten thousand
ten thousand
million
million
0
million
million
hundred
900
Exercise:
Problem: 909
hundred
Exercise:
Problem: 981
hundred
Solution:
hundred
1,000
Exercise:
thousand
1,000
thousand
thousand
thousand
1,000
ten thousand
0
ten thousand
ten thousand
ten thousand
0
million
0
million
million
million
0
Problem: 965
hundred thousand
Exercise:
Problem: 551,061,285
hundred thousand
Solution:
hundred thousand
551,061,300 551,061,000
Exercise:
Problem: 23,047,991,521
ten thousand
ten thousand
ten thousand
951,060,000
million
million
million
551,000,000
hundred thousand ten thousand
Exercise:
Problem: 106,999,413,206
Hundred thousand ten thousand
Solution:
hundred thousand ten thousand
106,999,413,200 106,999,413,000 106,999,410,000
Exercise:
Problem: 5,000,000
hundred thousand ten thousand
Exercise:
million
million
million
106,999,000,000
million
Problem: 8,006,001
hundred thousand ten thousand
Solution:
Hundred Thousand ten thousand
8,006,000 8,006,000 8,010,000
Exercise:
Problem: 94,312
hundred thousand ten thousand
Exercise:
Problem: 33,486
million
Million
8,000,000
million
hundred thousand
Solution:
hundred thousand
33,500 33,000
Exercise:
Problem: 560,669
hundred thousand
Exercise:
Problem: 388,551
hundred thousand
Solution:
ten thousand
ten thousand
30,000
ten thousand
ten thousand
million
million
0
million
million
hundred
388,600
Exercise:
Problem: 4,752
hundred
Exercise:
Problem: 8,209
hundred
Solution:
hundred
8,200
Exercise:
thousand
389,000
thousand
thousand
thousand
8,000
ten thousand
390,000
ten thousand
ten thousand
ten thousand
10,000
million
0
million
million
million
0
Problem:
In 1950, there were 5,796 cases of diphtheria reported in the United States. Round to
the nearest hundred.
Exercise:
Problem:
In 1979, 19,309,000 people in the United States received federal food stamps. Round
to the nearest ten thousand.
Solution:
19,310,000
Exercise:
Problem:
In 1980, there were 1,105,000 people between 30 and 34 years old enrolled in school.
Round to the nearest million.
Exercise:
Problem:
In 1980, there were 29,100,000 reports of aggravated assaults in the United States.
Round to the nearest million.
Solution:
29,000,000
For the following problems, round the numbers to the position you think is most
reasonable for the situation.
Exercise:
Problem:
In 1980, for a city of one million or more, the average annual salary of police and
firefighters was $16,096.
Exercise:
Problem:
The average percentage of possible sunshine in San Francisco, California, in June is
73%.
Solution:
70% or 75%
Exercise:
Problem:
In 1980, in the state of Connecticut, $3,777,000,000 in defense contract payroll was
awarded.
Exercise:
Problem:
In 1980, the federal government paid $5,463,000,000 to Viet Nam veterans and
dependants.
Solution:
$5,500,000,000
Exercise:
Problem: In 1980, there were 3,377,000 salespeople employed in the United States.
Exercise:
Problem:
In 1948, in New Hampshire, 231,000 popular votes were cast for the president.
Solution:
230,000
Exercise:
Problem: In 1970, the world production of cigarettes was 2,688,000,000,000.
Exercise:
Problem:
In 1979, the total number of motor vehicle registrations in Florida was 5,395,000.
Solution:
5,400,000
Exercise:
Problem: In 1980, there were 1,302,000 registered nurses the United States.
Exercises for Review
Exercise:
Problem:
([link]) There is a term that describes the visual displaying of a number. What is the
term?
Solution:
graphing
Exercise:
Problem: ((link]) What is the value of 5 in 26,518,206?
Exercise:
Problem: ((link]) Write 42,109 as you would read it.
Solution:
Forty-two thousand, one hundred nine
Exercise:
Problem: ([link]) Write "six hundred twelve" using digits.
Exercise:
Problem: ([link]) Write "four billion eight" using digits.
Solution:
4,000,000,008
Addition of Whole Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to add whole numbers. By
the end of this module, students should be able to understand the addition
process, add whole numbers, and use the calculator to add one whole
number to another.
Section Overview
e Addition
e Addition Visualized on the Number Line
e The Addition Process
e Addition Involving Carrying
e Calculators
Addition
Suppose we have two collections of objects that we combine together to
form a third collection. For example,
: : ' aE P BEES
is combined with = to yield Tr
a @
a G
We are combining a collection of four objects with a collection of three
objects to obtain a collection of seven objects.
Addition
The process of combining two or more objects (real or intuitive) to form a
third, the total, is called addition.
In addition, the numbers being added are called addends or terms, and the
total is called the sum. The plus symbol (+) is used to indicate addition,
and the equal symbol (=) is used to represent the word "equal." For
example, 4+ 3 = 7 means "four added to three equals seven."
Addition Visualized on the Number Line
Addition is easily visualized on the number line. Let's visualize the addition
of 4 and 3 using the number line.
To find 4 + 3,
1. Start at 0.
2. Move to the right 4 units. We are now located at 4.
3. From 4, move to the right 3 units. We are now located at 7.
Thus, 4+ 3 = 7.
The Addition Process
We'll study the process of addition by considering the sum of 25 and 43.
25
443 means
2 tens + 5 ones
+4 tens + 3 ones
6 tens + 8 ones
We write this as 68.
We can suggest the following procedure for adding whole numbers using
this example.
Example:
The Process of Adding Whole Numbers
To add whole numbers,
The process:
1. Write the numbers vertically, placing corresponding positions in the
same column.
25
+43
2. Add the digits in each column. Start at the right (in the ones position)
and move to the left, placing the sum at the bottom.
25
+43
68
Note: Confusion and incorrect sums can occur when the numbers are not
aligned in columns properly. Avoid writing such additions as
25
+43
25
+43
Sample Set A
Example:
Add 276 and 103.
276 64+3=49.
uate Oke
379 2+1=3.
Example:
Add 1459 and 130
1459 d ay ‘ > ‘
+130 | ae
1589 ex
jee (es ib
In each of these examples, each individual sum does not exceed 9. We will
examine individual sums that exceed 9 in the next section.
Practice Set A
Perform each addition. Show the expanded form in problems 1 and 2.
Exercise:
Problem: Add 63 and 25.
Solution:
88
6 tens + 3 ones
+2 tens + 5 ones
8 tens + 8 ones
Exercise:
Problem: Add 4,026 and 1,501.
Solution:
D.o27
4 thousands + 0 hundreds + 2 tens + 6 ones
+1 thousand +5 hundreds + 0 tens + 1 one
5 thousands + 5 hundreds + 2 tens + 7 ones
Exercise:
Problem: Add 231,045 and 36,121.
Solution:
267,166
Addition Involving Carrying
It often happens in addition that the sum of the digits in a column will
exceed 9. This happens when we add 18 and 34. We show this in expanded
form as follows.
a This sum exceeds 9.
18=1ten + 8ones 12 ones
+34=3tens+ 4 ones a ,
4 tens + 12 ones = 4 tens + 1 ten + 2 ones
\ ee tte
= 5 tens + 2 ones
= 52
Notice that when we add the 8 ones to the 4 ones we get 12 ones. We then
convert the 12 ones to 1 ten and 2 ones. In vertical addition, we show this
conversion by carrying the ten to the tens column. We write a 1 at the top
of the tens column to indicate the carry. This same example is shown in a
shorter form as follows:
8 + 4 = 12 Write 2, carry 1 ten to the top of the next column to the left.
Sample Set B
Perform the following additions. Use the process of carrying when needed.
Example:
Add 1875 and 358.
111
1875
+ 358
2233
oo Write 3, carry 1 ten.
14+7+5=13 °&#£2Write 3, carry 1 hundred.
14+8+3=12 #£zWrite 2, carry 1 thousand.
ie al eae
The sum is 2233.
Example:
Add 89,208 and 4,946.
11 1
89,208
+ 4,946
94,154
8+-6=14 Write 4, carry 1 ten.
14+0+4=5 Write the 5 (nothing to carry).
2+9=11 Write 1, carry one thousand.
14+9+4=14 °#£2Write 4, carry one ten thousand.
i 3= 9
The sum is 94,154.
Example:
Add 38 and 95.
11
38
+ 95
133
chap is) Write 3, carry 1 ten.
14+3+9=13 °&#®4Write 3, carry 1 hundred.
eel
As you proceed with the addition, it is a good idea to keep in mind what is
actually happening.
38 means 3 tens + 8 ones
+96 + 9tens + 5 ones
—_ 12 tens +13 ones
= 12 tens +1 ten+ 3 ones
= 13 tens + 3 ones
= 1 hundred+ 3 tena + 3 ones
= 133
The sum is 133.
Example:
Find the sum 2648, 1359, and 861.
111
2648
1359
+ 861
4868
Ore) lao Write 8, carry 1 ten.
14+4+5+6=16 }#£2Write 6, carry 1 hundred.
14+6+3+8=18 }#£2Write 8, carry 1 thousand.
1+2+4+1=4
The sum is 4,868.
Numbers other than 1 can be carried as illustrated in [link].
Example:
Find the sum of the following numbers.
132 1
878016
9905
38951
+ 56817
983689
OO le et Write 9, carry the 1.
Boe Ors ool Write 8.
ee) ola ae Write 6, carry the 2.
2+8+9+8+6=33 #£Write 3, carry the 3.
Soop hae ea By lis) Write 8, carry the 1.
fois Write 9.
The sum is 983,689.
Example:
The number of students enrolled at Riemann College in the years 1984,
1985, 1986, and 1987 was 10,406, 9,289, 10,108, and 11,412, respectively.
What was the total number of students enrolled at Riemann College in the
years 1985, 1986, and 1987?
We can determine the total number of students enrolled by adding 9,289,
10,108, and 11,412, the number of students enrolled in the years 1985,
1986, and 1987.
1 ll
9,289
10,108
+11,412
30,809
The total number of students enrolled at Riemann College in the years
1985, 1986, and 1987 was 30,809.
Practice Set B
Perform each addition. For the next three problems, show the expanded
form.
Exercise:
Problem: Add 58 and 29.
Solution:
87
5 tens + 8 ones
+2tens+ 9 ones
7 tens + 17 ones
— Ttens + 1ten + 7ones
= 8tens + 7ones
ae oi
Exercise:
Problem: Add 476 and 85.
Solution:
561
4 hundreds + 7tens+ 6 ones
+ 8tens+ 5 ones
4 hundreds + 15 tens + 11 ones
= 4 hundreds + 15 tens + 1 ten+ 1 one
= 4 hundreds + 16 tens + 1 one
= 4 hundreds + 1 hundred + 6 tens + 1 one
= 5 hundreds + 6 tens + 1 one
= 561
Exercise:
Problem: Add 27 and 88.
Solution:
115
2tens+ 7 ones
+ 8tens+ 8 ones
10 tens + 15 ones
= 10 tens + 1 ten + 5 ones
= 11 tens + 5 ones
= 1 hundred + 1 ten + 5 ones
= TED
Exercise:
Problem: Add 67,898 and 85,627.
Solution:
153,020
For the next three problems, find the sums.
Exercise:
57
Problem: 26
84
Solution:
167
Exercise:
847
Problem: 825
796
Solution:
2,468
Exercise:
16,945
8,472
Problem: 387,721
21,059
629
Solution:
434,826
Calculators
Calculators provide a very simple and quick way to find sums of whole
numbers. For the two problems in Sample Set C, assume the use of a
calculator that does not require the use of an ENTER key (such as many
Hewlett-Packard calculators).
Sample Set C
Use a calculator to find each sum.
Example:
34 + 21
Type
Press
Type
Press
The sum is 55.
Example:
34
106 + 85 4
Type
Press
Type
322 + 406
106
85
Display Reads
34
34
21
eye)
Display
Reads
The calculator keeps a
Pee running subtotal
106
85
Press = 191 - 106+ 85
Type B22 B22
Press 45 513 -— 191 + 322
Type 406 406
Press = 919 - 513 + 406
The sum is 919.
Practice Set C
Use a calculator to find the following sums.
Exercise:
Problem: 62 + 81+ 12
Solution:
155
Exercise:
Problem: 9,261 + 8,543 + 884 + 1,062
Solution:
19,750
Exercise:
Problem: 10,221 + 9,016 + 11,445
Solution:
30,682
Exercises
For the following problems, perform the additions. If you can, check each
sum with a calculator.
Exercise:
Problem:14 + 5
Solution:
19
Exercise:
Problem: 12 + 7
Exercise:
Problem: 46 + 2
Solution:
48
Exercise:
Problem: 83 + 16
Exercise:
Problem: 77 + 21
Solution:
98
Exercise:
Problem
Exercise:
321
“SAD
916
Problem:
“62
Solution:
978
Exercise:
Problem
Exercise:
Problem
104
"4561
265
"4103
Solution:
368
Exercise:
Problem
Exercise:
Problem
> 502 + 237
: 8,521 + 4,256
Solution:
12777
Exercise:
oe 16,408
roblem: + 3,101
Exercise:
pebi 16,515
roblem: 442,223
Solution:
58,738
Exercise:
Problem: 616,702 + 101,161
Exercise:
Problem: 43,156,219 + 2,013,520
Solution:
45,169,739
Exercise:
Problem: 17 + 6
Exercise:
Problem: 25 + 8
Solution
33
Exercise:
Problem
Exercise:
Problem
84
ae ak
+ 6
Solution:
81
Exercise:
Problem
Exercise:
Problem
: 36+ 48
:74+17
Solution:
a1
Exercise:
Problem
Exercise:
Problem
: 486 + 58
: 743 + 66
Solution:
809
Exercise:
Problem
Exercise:
Problem
: 381 + 88
687
p75
Solution:
862
Exercise:
Problem
Exercise:
Problem
931
"4853
: 1,428 + 893
Solution:
2,321
Exercise:
Problem
Exercise:
: 12,898 + 11,925
631,464
Problem:
+509,740
Solution:
1,141,204
Exercise:
Beeb 805,996
ro om) 3 98,516
Exercise:
38,428,106
Problem:
+522,936,005
Solution:
961,364,111
Exercise:
Problem: 5,288,423,100 + 16,934,785 ,995
Exercise:
Problem: 98,876,678,521,402 + 843,425,685,685,658
Solution:
942 ,302,364,207,060
Exercise:
Problem: 41 + 61+ 85 + 62
Exercise:
Problem: 21 + 85+ 104+9-+415
Solution:
234
Exercise:
116
27
Problem: 110
110
+ 8
Exercise:
75,206
Problem: 4,152
+16,007
Solution:
95,365
Exercise:
8,226
143
92,015
Problem:
8
487,553
5,218
Exercise:
50,006
1,005
100,300
20,008
1,000,009
800,800
Problem:
Solution:
1,972,128
Exercise:
616
42,018
1,687
225
Problem: 8,623,418
12,506,508
19
2,121
195,643
For the following problems, perform the additions and round to the nearest
hundred.
Exercise:
problem: 1488
robiem: 2,183
Solution:
3,700
Exercise:
Problem:
Exercise:
Problem:
Solution:
3,101,500
Exercise:
Problem:
Exercise:
Problem:
Solution:
100
Exercise:
Problem:
928,725
15,685
82,006
3,019,528
18,621
5,059
92
48
16
37
Exercise:
Pal
Problem:
16
Solution:
0
Exercise:
11172
Problem: 22,749
12,248
Exercise:
240
280
Problem:
210
310
Solution:
1,000
Exercise:
9,573
Problem: 101,279
122,581
For the next five problems, replace the letter m with the whole number that
will make the addition true.
Exercise:
Problem:
Solution:
is)
Exercise:
Problem:
Exercise:
Problem:
Solution:
19
Exercise:
Problem:
Exercise:
62
+ om
67
1,893
Problem: ++ m
1,981
Solution:
88
Exercise:
Problem:
The number of nursing and related care facilities in the United States
in 1971 was 22,004. In 1978, the number was 18,722. What was the
total number of facilities for both 1971 and 1978?
Exercise:
Problem:
The number of persons on food stamps in 1975, 1979, and 1980 was
19,179,000, 19,309,000, and 22,023,000, respectively. What was the
total number of people on food stamps for the years 1975, 1979, and
1980?
Solution:
60,511,000
Exercise:
Problem:
The enrollment in public and nonpublic schools in the years 1965,
1970, 1975, and 1984 was 54,394,000, 59,899,000, 61,063,000, and
55,122,000, respectively. What was the total enrollment for those
years?
Exercise:
Problem:
The area of New England is 3,618,770 square miles. The area of the
Mountain states is 863,563 square miles. The area of the South
Atlantic is 278,926 square miles. The area of the Pacific states is
921,392 square miles. What is the total area of these regions?
Solution:
5,682,651 square miles
Exercise:
Problem:
In 1960, the IRS received 1,188,000 corporate income tax returns. In
1965, 1,490,000 returns were received. In 1970, 1,747,000 returns
were received. In 1972 —1977, 1,890,000; 1,981,000; 2,043,000;
2,100,000; 2,159,000; and 2,329,000 returns were received,
respectively. What was the total number of corporate tax returns
received by the IRS during the years 1960, 1965, 1970, 1972 —1977?
Exercise:
Problem: Find the total number of scientists employed in 1974.
EMPLOYMENT STATUS OF MATHEMATICAL
SCIENTISTS — 1974
Solution:
1,190,000
Exercise:
Problem:
Find the total number of sales for space vehicle systems for the years
1965-1980.
SALES FOR SPACE VEHICLE SYSTEMS,
1965-1980
4 1,750,000,000}- ———
3 1,400,000,000
i
1965 1970 1971 1972 1973 1974 1975 1976
Year
Exercise:
Problem: Find the total baseball attendance for the years 1960-1980.
BASEBALL ATTENDANCE 1960-1980
1965 1970 1975 1977
Year
Solution:
271,564,000
Exercise:
Problem:
Find the number of prosecutions of federal officials for 1970-1980.
PROSECUTIONS OF FEDERAL OFFICIALS 1970-1980
Number of prosecutions
g 8
1971 1972 1973 1974 1975 1976 1977 1978 1979 1980
Year
For the following problems, try to add the numbers mentally.
Exercise:
Problem:
N W Oot Ol
Solution:
20
Exercise:
Problem:
Exercise:
Problem:
Solution:
23
Exercise:
Problem:
Exercise:
= & bo CO
bo Oot CO FR ©
“Iw OC Ot ND Ol
Problem:
Solution:
40
Exercise:
Problem:
Exercise:
Problem:
Solution:
50
Exercise:
Problem:
Exercise:
Pont nrewo er OD
20
30
15
39
16
14
|
Problem:
27
Solution:
50
Exercise:
82
Problem:
18
Exercise:
36
Problem:
14
Solution:
50
Exercises for Review
Exercise:
Problem:
({link]) Each period of numbers has its own name. From right to left,
what is the name of the fourth period?
Exercise:
Problem:
({link]) In the number 610,467, how many thousands are there?
Solution:
0
Exercise:
Problem
Exercise:
Problem
: (Llink]) Write 8,840 as you would read it.
: (Llink]) Round 6,842 to the nearest hundred.
Solution:
6,800
Exercise:
Problem
: (Llink]) Round 431,046 to the nearest million.
Subtraction of Whole Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to subtract whole numbers.
By the end of this module, students should be able to understand the
subtraction process, subtract whole numbers, and use a calculator to
subtract one whole number from another whole number.
Section Overview
e Subtraction
e Subtraction as the Opposite of Addition
e The Subtraction Process
e Subtraction Involving Borrowing
e Borrowing From Zero
e Calculators
Subtraction
Subtraction
Subtraction is the process of determining the remainder when part of the
total is removed.
Suppose the sum of two whole numbers is 11, and from 11 we remove 4.
Using the number line to help our visualization, we see that if we are
located at 11 and move 4 units to the left, and thus remove 4 units, we will
be located at 7. Thus, 7 units remain when we remove 4 units from 11 units.
The Minus Symbol
The minus symbol (-) is used to indicate subtraction. For example, 11 — 4
indicates that 4 is to be subtracted from 11.
Minuend
The number immediately in front of or the minus symbol is called the
minuend, and it represents the original number of units.
Subtrahend
The number immediately following or below the minus symbol is called the
subtrahend, and it represents the number of units to be removed.
Difference
The result of the subtraction is called the difference of the two numbers.
For example, in 11 — 4 = 7, 11 is the minuend, 4 is the subtrahend, and 7 is
the difference.
Subtraction as the Opposite of Addition
Subtraction can be thought of as the opposite of addition. We show this in
the problems in Sample Set A.
Sample Set A
Example:
8—5=3sincee3+5=8.
Example:
9—3=6since6+3=9.
Practice Set A
Complete the following statements.
Exercise:
Problem:7 — 5 = since +5 = 7.
Solution:
7—5=2since2+5=—7
Exercise:
Problem:9 — 1 = since +1 = 9.
Solution:
9—1= 8since8+1=9
Exercise:
Problem:17 — 8 = since +8 = 17.
Solution:
17 — 8= 9since 9+- 8 = 17
The Subtraction Process
We'll study the process of the subtraction of two whole numbers by
considering the difference between 48 and 35.
48 means 4 tens + 8 ones
— 35 —3 tens — 5 ones
lten +3 ones
which we write as 13.
Example:
The Process of Subtracting Whole Numbers
To subtract two whole numbers,
The process
1. Write the numbers vertically, placing corresponding positions in the
same column.
48
—35
2. Subtract the digits in each column. Start at the right, in the ones
position, and move to the left, placing the difference at the bottom.
48
—30
13
Sample Set B
Perform the following subtractions.
Example:
215
—142
133
5-2=3.
7-—4=8.
2 esd,
Example:
46,042
— 1,031
45,011
2—1=1.
4—3=1.
O= 0:0:
6 A= 5.
4—0=4.,
Example:
Find the difference between 977 and 235.
Write the numbers vertically, placing the larger number on top. Line up the
columns properly.
977
—2395
5:
The difference between 977 and 235 is 742.
Example:
In Keys County in 1987, there were 809 cable television installations. In
Flags County in 1987, there were 1,159 cable television installations. How
many more cable television installations were there in Flags County than in
Keys County in 1987?
We need to determine the difference between 1,159 and 809.
There were 350 more cable television installations in Flags County than in
Keys County in 1987.
Practice Set B
Perform the following subtractions.
Exercise:
Probl 534
m:
Pawar caja
Solution:
331
Exercise:
eT 857
r m:
oblem: — 13
Solution:
814
Exercise:
—s 95,628
roblem: 34,510
Solution:
61,118
Exercise:
— 11,005
ro em: 1,005
Solution:
10,000
Exercise:
Problem: Find the difference between 88,526 and 26,412.
Solution:
62,114
In each of these problems, each bottom digit is less than the corresponding
top digit. This may not always be the case. We will examine the case where
the bottom digit is greater than the corresponding top digit in the next
section.
Subtraction Involving Borrowing
Minuend and Subtrahend
It often happens in the subtraction of two whole numbers that a digit in the
minuend (top number) will be less than the digit in the same position in the
subtrahend (bottom number). This happens when we subtract 27 from 84.
84
—27
We do not have a name for 4 — 7. We need to rename 84 in order to
continue. We'll do so as follows:
84 = 8 tens + 4 ones
—~ 27 = 2 tens + 7 ones
7 tens + 1 ten + 4 ones
2 tens + 7 ones
7 tens + 10 ones + 4 ones
2 tens + 7 ones
Our new name for 84 is 7 tens + 14 ones.
7 tens + 14 ones
2tens+ 7 ones
5 tens + 7 ones
=D1
Notice that we converted 8 tens to 7 tens + 1 ten, and then we converted the
1 ten to 10 ones. We then had 14 ones and were able to perform the
subtraction.
Borrowing
The process of borrowing (converting) is illustrated in the problems of
Sample Set C.
Sample Set C
Example:
714
i
=21
57
1. Borrow 1 ten from the 8 tens. This leaves 7 tens.
2. Convert the 1 ten to 10 ones.
3. Add 10 ones to 4 ones to get 14 ones.
Example:
517
$72
— 91
581
1. Borrow 1 hundred from the 6 hundreds. This leaves 5 hundreds.
2. Convert the 1 hundred to 10 tens.
3. Add 10 tens to 7 tens to get 17 tens.
Practice Set C
Perform the following subtractions. Show the expanded form for the first
three problems.
Exercise:
Problem:
roblem _ 95
Solution:
18, 5 tens +3 ones
— 3 tens + 5 ones
4 tens + 1 ten + 3 ones
- 3 tens + 5 ones
4 tens + 13 ones
— 8tens+ 5 ones
lten + 8ones
=18
Exercise:
Problem: _98
Solution:
48, 7 tens + 6 ones
2 tens + 8 ones
6 tens + 1 ten + 6 ones
— 2tens + 8 ones
6 tens + 16 ones
— 2tens+ 8ones
4tens+ 8ones
= 48
Exercise:
Probl isle
roblem: _ o.
Solution:
307, 8 hundreds + 7 tens + 2 ones
— 5 hundreds + 6 tens + 5 ones
8 hundreds + 6 tens + 1 ten + 2 ones
- 5 hundreds + 6 tens + 5 ones
8 hundreds + 6 tens + 12 ones
_ 5 hundreds + 6 tens + 5 ones
3 hundreds + 0 tens + 7 ones
= 307
Exercise:
Probl 441
roblem: 356
Solution:
85
Exercise:
Baa 775
m:
roblem: _ 66
Solution:
709
Exercise:
er 5,663
roblem: 2,559
Solution:
3,104
Borrowing More Than Once
Sometimes it is necessary to borrow more than once. This is shown in the
problems in [link].
Sample Set D
Perform the Subtractions. Borrowing more than once if necessary
Example:
513
$11
¢41
— 358
283
1. Borrow 1 ten from the 4 tens. This leaves 3 tens.
2. Convert the 1 ten to 10 ones.
3. Add 10 ones to 1 one to get 11 ones. We can now perform 11 — 8.
4. Borrow 1 hundred from the 6 hundreds. This leaves 5 hundreds.
5. Convert the 1 hundred to 10 tens.
6. Add 10 tens to 3 tens to get 13 tens.
7. Now 13 — 5 = 8.
8.5-—3=2.
Example:
12
4914
B34
—__85
449
1. Borrow 1 ten from the 3 tens. This leaves 2 tens.
2. Convert the 1 ten to 10 ones.
3. Add 10 ones to 4 ones to get 14 ones. We can now perform 14 — 5.
4. Borrow 1 hundred from the 5 hundreds. This leaves 4 hundreds.
5. Convert the 1 hundred to 10 tens.
6. Add 10 tens to 2 tens to get 12 tens. We can now perform 12 — 8 = 4.
7. Finally, 4 — 0 = 4.
Example:
71529
- 6952
After borrowing, we have
10
14
69412
71529
— 6952
64577
Practice Set D
Perform the following subtractions.
Exercise:
peeuieni 526
roblem: 358
Solution:
168
Exercise:
63,419
Problem: _ 7,779
Solution:
55,640
Exercise:
4,312
Problem: 3,123
Solution:
1,189
Borrowing from Zero
It often happens in a subtraction problem that we have to borrow from one
or more zeros. This occurs in problems such as
i 503
'— 37
and
5000
We'll examine each case.
Example:
Borrowing from a single zero.
Consider the problem 37
Since we do not have a name for 3 — 7, we must borrow from 0.
503 = 5 hundreds + 0 tens + 3 ones
— 37 3 tens + 7 ones
Since there are no tens to borrow, we must borrow 1 hundred. One hundred
= 10 tens.
4 hundreds + 10 tens + 3 ones
3 tens + 7 ones
We can now borrow 1 ten from 10 tens (leaving 9 tens). One ten = 10 ones
and 10 ones + 3 ones = 13 ones.
4 hundreds + 9 tens + 13 ones
38tens+ 7 ones
4 hundreds + 6 tens + 6 ones = 466
Now we can suggest the following method for borrowing from a single
zero.
Borrowing from a Single Zero
To borrow from a single zero,
1. Decrease the digit to the immediate left of zero by one.
2. Draw a line through the zero and make it a 10.
3. Proceed to subtract as usual.
Sample Set E
Example:
Perform this subtraction.
503
=o
The number 503 contains a single zero
1. The number to the immediate left of 0 is 5. Decrease 5 by 1.
5-1=4
410
$93
— oF
2. Draw a line through the zero and make it a 10.
3. Borrow from the 10 and proceed.
9
41013
303
— $7
466
1 ten + 10 ones
10 ones + 3 ones = 13 ones
Practice Set E
Perform each subtraction.
Exercise:
Probl se
roblem: _ 18
Solution:
888
Exercise:
poi 5102
roblem: _ 559
Solution:
4,543
Exercise:
Bi 9055
roblem: _ 336
Solution:
8,669
Example:
Borrowing from a group of zeros
5000
Site
In this case, we have a group of zeros.
Consider the problem
5000 = 5 thousands + 0 hundred + 0 tens + 0 ones
— 37= 3 tens + 7 ones
Since we cannot borrow any tens or hundreds, we must borrow 1 thousand.
One thousand = 10 hundreds.
4 thousands + 10 hundreds + 0 tens + 0 ones
3 tens + 7 ones
We can now borrow 1 hundred from 10 hundreds. One hundred = 10 tens.
4 thousands + 9 hundreds + 10 tens + 0 ones
3 tens + 7 ones
We can now borrow 1 ten from 10 tens. One ten = 10 ones.
4 thousands + 9 hundreds + 9 tens + 10 ones
3tens+ 7 ones
4 thousands + 9 hundreds + 6 tens + 3 ones = 4,963
From observations made in this procedure we can suggest the following
method for borrowing from a group of zeros.
Borrowing from a Group of zeros
To borrow from a group of zeros,
1. Decrease the digit to the immediate left of the group of zeros by one.
2. Draw a line through each zero in the group and make it a 9, except the
rightmost zero, make it 10.
3. Proceed to subtract as usual.
Sample Set F
Perform each subtraction.
Example:
40,000
Zo
The number 40,000 contains a group of zeros.
1. The number to the immediate left of the group is 4. Decrease 4 by 1.
4—1=3
2. Make each 0, except the rightmost one, 9. Make the rightmost 0 a 10.
39 9910
49,000
—- 125
3. Subtract as usual.
39 9910
49,000
—- 125
39,875
Example:
8,000,006
SAH
The number 8,000,006 contains a group of zeros.
1. The number to the immediate left of the group is 8. Decrease 8 by 1.
eh =i if
2. Make each zero, except the rightmost one, 9. Make the rightmost 0 a
10.
7999 910
$,900,006
— 41,107
3. To perform the subtraction, we’ll need to borrow from the ten.
9
7 999 91016
$,909,006
— 41,107
7,958,899
1 ten = 10 ones
10 ones + 6 ones = 16 ones
Practice Set F
Perform each subtraction.
Exercise:
Problem: eo
— 4,873
Solution:
16,134
Exercise:
10,004
Problem: — 5,165
Solution:
4,839
Exercise:
ar 16,000,000
Tr a, ASOIED
Solution:
15,789,940
Calculators
In practice, calculators are used to find the difference between two whole
numbers.
Sample Set G
Find the difference between 1006 and 284.
Display Reads
Type 1006 1006
Press — 1006
Type 284 284
Press = Joe
The difference between 1006 and 284 is 722.
(What happens if you type 284 first and then 1006? We'll study such
numbers in [link]Chapter 10.)
Practice Set G
Exercise:
Problem:
Use a calculator to find the difference between 7338 and 2809.
Solution:
4,529
Exercise:
Problem:
Use a calculator to find the difference between 31,060,001 and
8,591,774.
Solution:
22,468,227
Exercises
For the following problems, perform the subtractions. You may check each
difference with a calculator.
Exercise:
15
Problem:
— 8
Solution:
7
Exercise:
19
Problem:
— 8
Exercise:
11
Problem:
— 5
Solution:
6
Exercise:
14
Problem:
— 6
Exercise:
12
Problem: 9
Solution:
3
Exercise:
56
Problem:
Exercise:
—12
74
Problem:
—33
Solution:
41
Exercise:
80
Problem:
Exercise:
Problem
—61
_ 390
*—141
Solution:
209
Exercise:
Problem
Exercise:
_ 800
* —650
Problem
35,002
"14,001
Solution:
21,001
Exercise:
Problem
Exercise:
Problem
5,000,566
* 2 441 326
400,605
* 121,352
Solution:
279,293
Exercise:
46,400
Problem:
Exercise:
91D
77,893
Problem:
A21
Solution:
77,472
Exercise:
A2
Problem:
Exercise:
—18
dl
Problem:
=—2%
Solution:
24
Exercise:
622
Problem:
Exercise:
— 88
261
Problem:
— 73
Solution:
188
Exercise:
Problem
Exercise:
Problem
242
" —158
3,422
1045
Solution:
2377
Exercise:
Probl 5,965
roblem: 3,985
Exercise:
— 42,041
ro em: 15 355
Solution:
26,686
Exercise:
Probl 304,056
roblem: _ 20,008
Exercise:
Probl 64,000,002
Pen oss. “BRGOTAS
Solution:
63,143,259
Exercise:
4,109
Problem:
856
Exercise:
10,113
Problem:
2) 9.079
Solution:
8,034
Exercise:
605
Problem:
Exercise:
a as
59
Problem:
—26
Solution:
33
Exercise:
36,107
Problem:
Exercise:
Problem
V8 814
92.526,441,820
” 59,914,805,253
Solution:
32,611,636,567
Exercise:
peanienL 1,605
roblem: 981
Exercise:
aoe 30,000
roblem: 26,062
Solution:
3,938
Exercise:
Bsa 600
roblem: 916
Exercise:
pei 9,000,003
Fromme 796.048
Solution:
6:273,955
For the following problems, perform each subtraction.
Exercise:
Problem: Subtract 63 from 92.
Note: The word "from" means "beginning at." Thus, 63 from 92
means beginning at 92, or 92 — 63.
Exercise:
Problem: Subtract 35 from 86.
Solution:
51
Exercise:
Problem: Subtract 382 from 541.
Exercise:
Problem: Subtract 1,841 from 5,246.
Solution:
3,405
Exercise:
Problem: Subtract 26,082 from 35,040.
Exercise:
Problem: Find the difference between 47 and 21.
Solution:
26
Exercise:
Problem:
Exercise:
Problem:
Solution:
72,069
Exercise:
Problem:
Exercise:
Problem:
Solution:
B17
Exercise:
Problem:
Exercise:
Problem:
Solution:
29
Exercise:
Problem:
Find the difference between 1,005 and 314.
Find the difference between 72,085 and 16.
Find the difference between 7,214 and 2,049.
Find the difference between 56,108 and 52,911.
How much bigger is 92 than 47?
How much bigger is 114 than 85?
How much bigger is 3,006 than 1,918?
Exercise:
Problem: How much bigger is 11,201 than 816?
Solution:
10,385
Exercise:
Problem: How much bigger is 3,080,020 than 1,814,161?
Exercise:
Problem:
In Wichita, Kansas, the sun shines about 74% of the time in July and
about 59% of the time in November. How much more of the time (in
percent) does the sun shine in July than in November?
Solution:
15%
Exercise:
Problem:
The lowest temperature on record in Concord, New Hampshire in May
is 21°F, and in July it is 35°F. What is the difference in these lowest
temperatures?
Exercise:
Problem:
In 1980, there were 83,000 people arrested for prostitution and
commercialized vice and 11,330,000 people arrested for driving while
intoxicated. How many more people were arrested for drunk driving
than for prostitution?
Solution:
11,247,000
Exercise:
Problem:
In 1980, a person with a bachelor's degree in accounting received a
monthly salary offer of $1,293, and a person with a marketing degree a
monthly salary offer of $1,145. How much more was offered to the
person with an accounting degree than the person with a marketing
degree?
Exercise:
Problem:
In 1970, there were about 793 people per square mile living in Puerto
Rico, and 357 people per square mile living in Guam. How many more
people per square mile were there in Puerto Rico than Guam?
Solution:
436
Exercise:
Problem:
The 1980 population of Singapore was 2,414,000 and the 1980
population of Sri Lanka was 14,850,000. How many more people lived
in Sri Lanka than in Singapore in 1980?
Exercise:
Problem:
In 1977, there were 7,234,000 hospitals in the United States and
64,421,000 in Mainland China. How many more hospitals were there
in Mainland China than in the United States in 1977?
Solution:
57,187,000
Exercise:
Problem:
In 1978, there were 3,095,000 telephones in use in Poland and
4,292,000 in Switzerland. How many more telephones were in use in
Switzerland than in Poland in 1978?
For the following problems, use the corresponding graphs to solve the
problems.
Exercise:
Problem:
How many more life scientists were there in 1974 than
mathematicians? ({link])
Solution:
165,000
Exercise:
Problem:
How many more social, psychological, mathematical, and
environmental scientists were there than life, physical, and computer
scientists? ([link])
EMPLOYMENT STATUS OF SCIENTISTS— 1974
Exercise:
Problem:
How many more prosecutions were there in 1978 than in 1974?
({link])
Solution:
74
Exercise:
Problem:
How many more prosecutions were there in 1976-1980 than in 1970-
1975? ([link])
PROSECUTIONS OF FEDERAL OFFICIALS 1970-1980
1971 1972 1973 1974 1975 1976 1977 1978 1979 1980
Year
Exercise:
Problem:
How many more dry holes were drilled in 1960 than in 1975? ([link])
Solution:
4,547
Exercise:
Problem:
How many more dry holes were drilled in 1960, 1965, and 1970 than
in 1975, 1978 and 1979? ([link])
OIL WELLS—DRY HOLES DRILLED 1960-1979
Year
For the following problems, replace the L] with the whole number that will
make the subtraction true.
Exercise:
14
Problem: — |]
3
Solution:
sla
Exercise:
Zi.
Problem: — 1]
14
Exercise:
35
Problem: — |
25
Solution:
10
Exercise:
16
Problem: — |]
9
Exercise:
28
Problem: — |]
16
Solution:
12
For the following problems, find the solutions.
Exercise:
Problem: Subtract 42 from the sum of 16 and 56.
Exercise:
Problem: Subtract 105 from the sum of 92 and 89.
Solution:
76
Exercise:
Problem: Subtract 1,127 from the sum of 2,161 and 387.
Exercise:
Problem: Subtract 37 from the difference between 263 and 175.
Solution:
pi
Exercise:
Problem: Subtract 1,109 from the difference between 3,046 and 920.
Exercise:
Problem:
Add the difference between 63 and 47 to the difference between 55
and 11.
Solution:
60
Exercise:
Problem:
Add the difference between 815 and 298 to the difference between
2,204 and 1,016.
Exercise:
Problem:
Subtract the difference between 78 and 43 from the sum of 111 and 89.
Solution:
165
Exercise:
Problem:
Subtract the difference between 18 and 7 from the sum of the
differences between 42 and 13, and 81 and 16.
Exercise:
Problem:
Find the difference between the differences of 343 and 96, and 521 and
488.
Solution:
214
Exercises for Review
Exercise:
Problem:
({link]) In the number 21,206, how many hundreds are there?
Exercise:
Problem:
({link]) Write a three-digit number that has a zero in the ones position.
Solution:
330 (answers may vary)
Exercise:
Problem: ({link]) How many three-digit whole numbers are there?
Exercise:
Problem: ({link]) Round 26,524,016 to the nearest million.
Solution:
27,000,000
Exercise:
Problem: ((link]) Find the sum of 846 + 221 + 116.
Properties of Addition
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses properties of addition. By the end
of the module students should be able to understand the commutative and
associative properties of addition and understand why 0 is the additive
identity.
Section Overview
e The Commutative Property of Addition
e The Associative Property of Addition
e The Additive Identity
We now consider three simple but very important properties of addition.
The Commutative Property of Addition
Commutative Property of Addition
If two whole numbers are added in any order, the sum will not change.
Sample Set A
Example:
Add the whole numbers
The numbers 8 and 5 can be added in any order. Regardless of the order
they are added, the sum is 13.
Practice Set A
Exercise:
Problem:
Use the commutative property of addition to find the sum of 12 and 41
in two different ways.
Solution:
12 + 41 = 53 and 41 + 12 = 53
Exercise:
Problem: Add the whole numbers
837
1,958
Solution:
837 + 1,958 = 2,795 and 1,958 + 837 = 2,795
The Associative Property of Addition
Associative Property of Addition
If three whole numbers are to be added, the sum will be the same if the first
two are added first, then that sum is added to the third, or, the second two
are added first, and that sum is added to the first.
Using Parentheses
It is a common mathematical practice to use parentheses to show which
pair of numbers we wish to combine first.
Sample Set B
Example:
Add the whole numbers.
43 and 16 are associated.
43 (43 + 16) + 27 = 59 + 27 = 86.
16 43 + (16 + 27) = 43 + 43 = 86.
27 {16 and 27 are associated.
Practice Set B
Exercise:
Problem:
Use the associative property of addition to add the following whole
numbers two different ways.
Solution:
(17 + 32) + 25 = 49 + 25 = 74 and
17 + (32 +25) =17+457= 74
Exercise:
Problem:
1,629
806
429
Solution:
(1,629 + 806) + 429 = 2,435 + 429 = 2,864
1,629 + (806 + 429) = 1,629 + 1,235 = 2,864
The Additive Identity
0 Is the Additive Identity
The whole number 0 is called the additive identity, since when it is added
to any whole number, the sum is identical to that whole number.
Sample Set C
Example:
Add the whole numbers.
Za NaeAI f—s 748)
Oo 29 — 29
Zero added to 29 does not change the identity of 29.
Practice Set C
Add the following whole numbers.
Exercise:
Problem:
Solution:
8
Exercise:
Problem:
Solution:
3)
Suppose we let the letter x represent a choice for some whole number. For
the first two problems, find the sums. For the third problem, find the sum
provided we now know that x represents the whole number 17.
Exercise:
Problem:
Solution:
x
Exercise:
Problem:
0
x
Solution:
x
Exercise:
Problem:
Solution:
L7
Exercises
For the following problems, add the numbers in two ways.
Exercise:
Problem:
Solution:
a7
Exercise:
Problem:
a6
12
Exercise:
Problem:
Solution:
45
Exercise:
Problem:
lit
Exercise:
Problem:
Solution:
568
Exercise:
Problem:
Exercise:
Problem:
Td, 205
49,118
Solution:
122.323
Exercise:
Problem:
Exercise:
Problem:
Solution:
45
Exercise:
Problem:
Exercise:
Problem:
Solution:
100
Exercise:
Problem:
Exercise:
Problem:
Solution:
556
Exercise:
Problem:
1019
11
5a
Exercise:
Problem:
Solution:
43,461
For the following problems, show that the pairs of quantities yield the same
sum.
Exercise:
Problem: (11 + 27) + 9 and 11 + (27+ 9)
Exercise:
Problem: (80 + 52) + 6 and 80 + (52 + 6)
Solution:
132 + 6 =80 + 58 = 138
Exercise:
Problem: (114 + 226) + 108 and 114 + (226 + 108)
Exercise:
Problem: (731 + 256) + 171 and 731 + (256 + 171)
Solution:
987 + 171 =731 + 427 = 1,158
Exercise:
Problem:
The fact that (a first number + a second number) + third number = a
first number + (a second number + a third number) is an example of
the property of addition.
Exercise:
Problem:
The fact that 0 + any number = that particular number is an example of
the property of addition.
Solution:
Identity
Exercise:
Problem:
The fact that a first number + a second number = a second number + a
first number is an example of the property of addition.
Exercise:
Problem:
Use the numbers 15 and 8 to illustrate the commutative property of
addition.
Solution:
15+8=8+4+15 = 23
Exercise:
Problem:
Use the numbers 6, 5, and 11 to illustrate the associative property of
addition.
Exercise:
Problem:
The number zero is called the additive identity. Why is the term
identity so appropriate?
Solution:
...because its partner in addition remains identically the same after that
addition
Exercises for Review
Exercise:
Problem: ({link]) How many hundreds in 46,581?
Exercise:
Problem: ({link]) Write 2,218 as you would read it.
Solution:
Two thousand, two hundred eighteen.
Exercise:
Problem: ({link]) Round 506,207 to the nearest thousand.
Exercise:
482
Problem: ([link]) Find the sum of + 68
Solution:
550
Exercise:
3,318
Problem: ({link]) Find the difference: 499
Summary of Key Concepts
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module reviews the key concepts from the chapter
Addition and Subtraction of Whole Numbers.
Summary of Key Concepts
Number / Numeral ({link])
A number is a concept. It exists only in the mind. A numeral is a symbol
that represents a number. It is customary not to distinguish between the two
(but we should remain aware of the difference).
Hindu-Arabic Numeration System ([link])
In our society, we use the Hindu-Arabic numeration system. It was
invented by the Hindus shortly before the third century and popularized by
the Arabs about a thousand years later.
Digits ([link])
The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are called digits.
Base Ten Positional System ((link])
The Hindu-Arabic numeration system is a positional number system with
base ten. Each position has value that is ten times the value of the position
to its right.
Commas / Periods ((link])
Commas are used to separate digits into groups of three. Each group of
three is called a period. Each period has a name. From right to left, they are
ones, thousands, millions, billions, etc.
Whole Numbers ((link])
A whole number is any number that is formed using only the digits (0, 1,
255, 4,0; 0, 7505 9).
Number Line ([link])
The number line allows us to visually display the whole numbers.
Graphing ([link])
Graphing a whole number is a term used for visually displaying the whole
number. The graph of 4 appears below.
Reading Whole Numbers ([link])
To express a whole number as a verbal phrase:
1. Begin at the right and, working right to left, separate the number into
distinct periods by inserting commas every three digits.
2. Begin at the left, and read each period individually.
Writing Whole Numbers ([link])
To rename a number that is expressed in words to a number expressed in
digits:
1. Notice that a number expressed as a verbal phrase will have its periods
set off by commas.
2. Start at the beginning of the sentence, and write each period of
numbers individually.
3. Use commas to separate periods, and combine the periods to form one
number.
Rounding ((link])
Rounding is the process of approximating the number of a group of objects
by mentally "seeing" the collection as occurring in groups of tens,
hundreds, thousands, etc.
Addition ({link])
Addition is the process of combining two or more objects (real or intuitive)
to form a new, third object, the total, or sum.
Addends / Sum ([link])
In addition, the numbers being added are called addends and the result, or
total, the sum.
Subtraction ({link])
Subtraction is the process of determining the remainder when part of the
total is removed.
Minuend / Subtrahend Difference (({link])
18—11=7
Be Jes
minuend subtrahend difference
Commutative Property of Addition ({link])
If two whole numbers are added in either of two orders, the sum will not
change.
34+5=5+4+3
Associative Property of Addition ([link])
If three whole numbers are to be added, the sum will be the same if the first
two are added and that sum is then added to the third, or if the second two
are added and the first is added to that sum.
(3+5)+2=3+4+(542)
Parentheses in Addition ({link])
Parentheses in addition indicate which numbers are to be added first.
Additive Identity ({link])
The whole number 0 is called the additive identity since, when it is added
to any particular whole number, the sum is identical to that whole number.
0+7=7
7+0=7
Exercise Supplement
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is an exercise supplement for the chapter
Addition and Subtraction of Whole Numbers and contains many exercise
problems. Odd problems are accompanied by solutions.
Exercise Supplement
For problems 1-35, find the sums and differences.
Exercise:
Bub 908
r :
oblem 4 99
Solution:
937
Exercise:
Brabiway: 529
roblem: 4161
Exercise:
TT 549
r :
oblem + 16
Solution:
565
Exercise:
726
Problem
* 4899
Exercise:
Problem
390
"+169
Solution:
aoe
Exercise:
Problem
Exercise:
Problem
166
* +660
391
"4951
Solution:
1,342
Exercise:
48
Problem:
Exercise:
+36
1,103
Problem:
"4 898
Solution:
2,001
Exercise:
1,642
Problem:
Exercise:
Problem
"4 899
807
"41,156
Solution:
1,963
Exercise:
Problem
Exercise:
Problem
80,349
"+ 2,679
70,070
"+ 9.386
Solution:
79,456
Exercise:
Problem
Exercise:
90,874
"4 2.945
Problem
A5,292
"451,661
Solution:
96,953
Exercise:
Problem
Exercise:
Problem
1,617
"454,923
702,607
“+ 89,217
Solution:
791,824
Exercise:
6,670,006
Problem:
Exercise:
Problem
+ 2.495
267
"+8 034
Solution:
8,301
Exercise:
Problem:
Exercise:
Problem:
Solution:
140,381
Exercise:
Problem:
Exercise:
Problem:
Solution:
76,224
Exercise:
Problem:
Exercise:
Problem:
7,007
+11,938
131,294
+ 9,087
5,292
+ 161
17,260
+58,964
7,006
—5,382
7,973
~3,018
Solution:
4,955
Exercise:
Problem
Exercise:
16,608
"— 1,660
209,527
Problem:
23,916
Solution:
185,611
Exercise:
Problem
Exercise:
Problem
_ 584
" —226
S316
"1.075
Solution:
2,238
Exercise:
Problem
458
" —122
Exercise:
1,007
Problem:
“ats B31
Solution:
1,336
Exercise:
Problem
Exercise:
16,082
"8.018
926
Problem:
— 48
Solution:
878
Exercise:
Problem
Exercise:
Problem
736
” 45,869
676,504
= BROTT
Solution:
618,227
For problems 36-39, add the numbers.
Exercise:
769
Problem: a
roblem: 598
746
Exercise:
554
Problem: 184
883
Solution:
1,621
Exercise:
30,188
79,731
pa 16,600
roblem: 66,085
39,169
95,170
Exercise:
2.129
6,190
17,044
Problem: 30,447
292
Al
428,458
Solution:
484,601
For problems 40-50, combine the numbers as indicated.
Exercise:
Problem: 2,957 + 9,006
Exercise:
Problem: 19,040 + 813
Solution:
Ae Metso.
Exercise:
Problem: 350,212 + 14,533
Exercise:
Problem: 970 + 702 + 22+ 8
Solution:
1,702
Exercise:
Problem:3,704 + 2,344 + 429 + 10,374 + 74
Exercise:
Problem: 874 + 845 + 295 — 900
Solution:
1,114
Exercise:
Problem: 904 + 910 — 881
Exercise:
Problem: 521 + 453 — 334 + 600
Solution:
1,300
Exercise:
Problem: 892 — 820 — 9
Exercise:
Problem: 159 + 4,085 — 918 — 608
Solution:
25718
Exercise:
Problem: 2,562 + 8,754 — 393 — 385 — 910
For problems 51-63, add and subtract as indicated.
Exercise:
Problem: Subtract 671 from 8,027.
Solution:
7300
Exercise:
Problem: Subtract 387 from 6,342.
Exercise:
Problem: Subtract 2,926 from 6,341.
Solution:
3,415
Exercise:
Problem: Subtract 4,355 from the sum of 74 and 7,319.
Exercise:
Problem: Subtract 325 from the sum of 7,188 and 4,964.
Solution:
11,827
Exercise:
Problem: Subtract 496 from the difference of 60,321 and 99.
Exercise:
Problem: Subtract 20,663 from the difference of 523,150 and 95,225.
Solution:
407,262
Exercise:
Problem:
Add the difference of 843 and 139 to the difference of 4,450 and 839.
Exercise:
Problem:
Add the difference of 997,468 and 292,513 to the difference of 22,140
and 8,617.
Solution:
718,478
Exercise:
Problem:
Subtract the difference of 8,412 and 576 from the sum of 22,140 and
8,617.
Exercise:
Problem:
Add the sum of 2,273, 3,304, 847, and 16 to the difference of 4,365
and 864.
Solution:
9,941
Exercise:
Problem:
Add the sum of 19,161, 201, 166,127, and 44 to the difference of the
sums of 161, 2,455, and 85, and 21, 26, 48, and 187.
Exercise:
Problem:
Is the sum of 626 and 1,242 the same as the sum of 1,242 and 626?
Justify your claim.
Solution:
626 + 1,242 = 1,242 + 626 = 1,868
Proficiency Exam
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is a proficiency exam to the chapter
Addition and Subtraction of Whole Numbers. Each problem is
accompanied with a reference link pointing back to the module that
discusses the type of problem demonstrated in the question. The problems
in this exam are accompanied by solutions.
Proficiency Exam
Exercise:
Problem: ({link]) What is the largest digit?
Solution:
9
Exercise:
Problem:
({link]) In the Hindu-Arabic number system, each period has three
values assigned to it. These values are the same for each period. From
right to left, what are they?
Solution:
ones, tens, hundreds
Exercise:
Problem:
({link]) In the number 42,826, how many hundreds are there?
Solution:
8
Exercise:
Problem: ({link]) Is there a largest whole number? If so, what is it?
Solution:
no
Exercise:
Problem:
({link]) Graph the following whole numbers on the number line: 2, 3,
Exercise:
Problem: ({link]) Write the number 63,425 as you would read it aloud.
Solution:
Sixty-three thousand, four hundred twenty-five
Exercise:
Problem:
({link]) Write the number eighteen million, three hundred fifty-nine
thousand, seventy-two.
Solution:
18,359,072
Exercise:
Problem: ({link]) Round 427 to the nearest hundred.
Solution:
400
Exercise:
Problem: ({link]) Round 18,995 to the nearest ten.
Solution:
19,000
Exercise:
Problem:
({link]) Round to the most reasonable digit: During a semester, a
mathematics instructor uses 487 pieces of chalk.
Solution:
500
For problems 11-17, find the sums and differences.
Exercise:
Probl link im
robiem.: ae) 4. A8
Solution:
675
Exercise:
Problem: ([link]) 3106 + 921
Solution:
4,027
Exercise:
152
Problem: ({link}) :
36
Solution:
188
Exercise:
5,189
Problem: ({link]) 4.122
+8,001
Solution:
23,501
Exercise:
Problem: (({link]) 21+ 16+ 42+ 11
Solution:
90
Exercise:
Problem: ((link]) 520 — 216
Solution:
304
Exercise:
Problem: ({link])
Solution:
70,125
Exercise:
80,001
— 9,878
Problem: ({link]) Subtract 425 from 816.
Solution:
hell
Exercise:
Problem: ({link]) Subtract 712 from the sum of 507 and 387.
Solution:
182
Exercise:
Problem:
({link]) Is the sum of 219 and 412 the same as the sum of 412 and 219°?
If so, what makes it so?
Solution:
Yes, commutative property of addition
Objectives
This module contains Chapter 2 of Fundamentals of Mathematics by Denny
Burzynski and Wade Ellis, Jr.
After completing this chapter, you should
Multiplication of Whole Numbers (({link])
e understand the process of multiplication
e be able to multiply whole numbers
e be able to simplify multiplications with numbers ending in zero
¢ be able to use a calculator to multiply one whole number by another
Concepts of Division of Whole Numbers ({link])
e understand the process of division
¢ understand division of a nonzero number into zero
¢ understand why division by zero is undefined
¢ be able to use a calculator to divide one whole number by another
Division of Whole Numbers ({link])
e be able to divide a whole number by a single or multiple digit divisor
e be able to interpret a calculator statement that a division results in a
remainder
Some Interesting Facts about Division ({link])
e be able to recognize a whole number that is divisible by 2, 3, 4, 5, 6, 8,
9, or 10
Properties of Multiplication ({link])
e understand and appreciate the commutative and associative properties
of multiplication
e understand why 1 is the multiplicative identity
Multiplication of Whole Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This
module discusses how to multiply whole numbers. By the end of the module students should be
able to understand the process of multiplication, multiply whole numbers, simplify
multiplications with numbers ending in zero, and use a calculator to multiply one whole number
by another.
Section Overview
e¢ Multiplication
The Multiplication Process With a Single Digit Multiplier
The Multiplication Process With a Multiple Digit Multiplier
e Multiplication With Numbers Ending in Zero
e Calculators
Multiplication
Multiplication is a description of repeated addition.
In the addition of
5+5+5
the number 5 is repeated 3 times. Therefore, we say we have three times five and describe it by
writing
3x5
Thus,
3xX5=5+5+4+5
Multiplicand
In a multiplication, the repeated addend (number being added) is called the multiplicand. In
3 x 5, the 5 is the multiplicand.
Multiplier
Also, in a multiplication, the number that records the number of times the multiplicand is used is
called the multiplier. In 3 x 5, the 3 is the multiplier.
Sample Set A
Express each repeated addition as a multiplication. In each case, specify the multiplier and the
multiplicand.
Example:
(eee pel tie pur We et Akai
6 x 7. Multiplier is 6. Multiplicand is 7.
Example:
18+18+18
3 xX 18. Multiplier is 3. Multiplicand is 18.
Practice Set A
Express each repeated addition as a multiplication. In each case, specify the multiplier and the
multiplicand.
Exercise:
Problem: 12 + 12+ 12+ 12
. Multiplier is . Multiplicand is .
Solution:
4 x 12. Multiplier is 4. Multiplicand is 12.
Exercise:
Problem: 36 + 36 + 36 + 36 + 36 + 36 + 36 + 36
. Multiplier is . Multiplicand is .
Solution:
8 x 36. Multiplier is 8. Multiplicand is 36.
Exercise:
Problem: 0+0+0+0+0
. Multiplier is . Multiplicand is .
Solution:
5 x 0. Multiplier is 5. Multiplicand is 0.
Exercise:
1847 + 1847 4+ ... + 1847
Problem:
12,000 times
. Multiplier is . Multiplicand is .
Solution:
12,000 x 1,847. Multiplier is 12,000. Multiplicand is 1,847.
Factors
In a multiplication, the numbers being multiplied are also called factors.
Products
The result of a multiplication is called the product. In 3 x 5 = 15, the 3 and 5 are not only called
the multiplier and multiplicand, but they are also called factors. The product is 15.
Indicators of Multiplication <,-,()
The multiplication symbol (x) is not the only symbol used to indicate multiplication. Other
symbols include the dot ( - ) and pairs of parentheses ( ). The expressions
3 x 5, 3-5, 3(5), (3)5, (3)(5)
all represent the same product.
The Multiplication Process With a Single Digit Multiplier
Since multiplication is repeated addition, we should not be surprised to notice that carrying can
occur. Carrying occurs when we find the product of 38 and 7:
First, we compute 7 x 8 = 56. Write the 6 in the ones column. Carry the 5. Then take
7 x 3 = 21. Add to 21 the 5 that was carried: 21 + 5 = 26. The product is 266.
Sample Set B
Find the following products.
Example:
o 7
x_3
192
eet iy)
Beco=6
Write the 2, carry the 1.
Add to 18 the 1 that was carried: 18 + 1 = 19.
The product is 192.
oo 25
Write the 0, carry the 3.
Add to 10 the 3 that was carried: 10 + 3 = 13. Write the 3, carry the 1.
Add to 25 the 1 that was carried: 25 + 1= 6.
The product is 2,630.
Example:
Write the 6, carry the 3.
Add to the 0 the 3 that was carried:0 + 3 = 3. Write the 3.
Write the 2, carry the 7.
Add to the 9 the 7 that was carried: 9 + 7 = 16.
Since there are no more multiplications to perform,write both the 1 and 6.
The product is 16,236.
Practice Set B
Find the following products.
Exercise:
37
Problem:
x 5
Solution:
185
Exercise:
78
Problem:
“x 8
Solution:
624
Exercise:
536
Problem:
Solution:
3,792
Exercise:
Problem:
40,019
x 8
Solution:
320,152
Exercise:
Problem
_ 301,599
“y 3
Solution:
904,797
The Multiplication Process With a Multiple Digit Multiplier
In a multiplication in which the multiplier is composed of two or more digits, the multiplication
must take place in parts. The process is as follows:
¢ Part 1 First Partial Product Multiply the multiplicand by the ones digit of the multiplier.
This product is called the first partial product.
¢ Part 2 Second Partial Product Multiply the multiplicand by the tens digit of the multiplier.
This product is called the second partial product. Since the tens digit is used as a factor, the
second partial product is written below the first partial product so that its rightmost digit
appears in the tens column.
e Part 3 If necessary, continue this way finding partial products. Write each one below the
previous one so that the rightmost digit appears in the column directly below the digit that
was used as a factor.
e Part 4 Total Product Add the partial products to obtain the total product.
Note:It may be necessary to carry when finding each partial product.
Sample Set C
Example:
Multiply 326 by 48.
e Part 1
24
326
x 48
2608 <— First partial product.
e Part 2
12
24
326
X_48
2608
1304 <— Second partial product.
e Part 3This step is unnecessary since all of the digits in the multiplier have been used.
e Part 4Add the partial products to obtain the total product.
12
24
326
x_48
2608
+1304
15648
<— Total product.
e The product is 15,648.
Example:
Multiply 5,369 by 842.
e Part 1
11
5369
X 842
10738
e Part 2
123
11
5369
X_842
10738
21476
e Part 3
257
123
11
5369
X_842
10738
21476
42952
4520698
<— First partial product.
«— Second partial product.
<— Third partial product.
<— Total product (Part 4).
e The product is 4,520,698.
Example:
Multiply 1,508 by 206.
e Part 1
34
1508
X 206
9048 #<— First partial product (in first column from the right).
e Part 2
364
1508
X_206
9048
Since 0 times 1508 is 0, the partial product will not change the identity of the total product
(which is obtained by addition). Go to the next partial product.
e Part 3
11
3.4
1508
x 206
9048
3016 == «— Third partial product (in third column from the right).
310648 <— Total product (Part 4).
e The product is 310,648
Practice Set C
Exercise:
Problem: Multiply 73 by 14.
Solution:
1,022
Exercise:
Problem: Multiply 86 by 52.
Solution:
4,472
Exercise:
Problem: Multiply 419 by 85.
Solution:
35,615
Exercise:
Problem: Multiply 2,376 by 613.
Solution:
1,456,488
Exercise:
Problem: Multiply 8,107 by 304.
Solution:
2,464,528
Exercise:
Problem: Multiply 66,260 by 1,008.
Solution:
66,790,080
Exercise:
Problem: Multiply 209 by 501.
Solution:
104,709
Exercise:
Problem: Multiply 24 by 10.
Solution:
240
Exercise:
Problem: Multiply 3,809 by 1,000.
Solution:
3,809,000
Exercise:
Problem: Multiply 813 by 10,000.
Solution:
8,130,000
Multiplications With Numbers Ending in Zero
Often, when performing a multiplication, one or both of the factors will end in zeros. Such
multiplications can be done quickly by aligning the numbers so that the rightmost nonzero digits
are in the same column.
Sample Set D
Perform the multiplication (49,000) (1,200).
(49,000)(1,200) = 49000
x 1200
Since 9 and 2 are the rightmost nonzero digits, put them in the same column.
49000
1200
Draw (perhaps mentally) a vertical line to separate the zeros from the nonzeros.
49/000
X 12/00
Multiply the numbers to the left of the vertical line as usual, then attach to the right end of this
product the total number of zeros.
Attach these 5 zeros to 588.
The product is 58,800,000
Practice Set D
Exercise:
Problem: Multiply 1,800 by 90.
Solution:
162,000
Exercise:
Problem: Multiply 420,000 by 300.
Solution:
126,000,000
Exercise:
Problem: Multiply 20,500,000 by 140,000.
Solution:
2,870,000,000,000
Calculators
Most multiplications are performed using a calculator.
Sample Set E
Example:
Multiply 75,891 by 263.
Display Reads
Type 75891 75891
Press x 75891
Type 263 263
Press - 19959333
The product is 19,959,333.
Example:
Multiply 4,510,000,000,000 by 1,700.
Display Reads
Type 451 451
Press x 451
Type 17 17
Press - 7667
The display now reads 7667. We'll have to add the zeros ourselves. There are a total of 12 zeros.
Attaching 12 zeros to 7667, we get 7,667,000,000,000,000.
The product is 7,667,000,000,000,000.
Example:
Multiply 57,847,298 by 38,976.
Display Reads
Type 57847298 57847298
Press x 57847298
Type 38976 38976
Press = 2.2546563 12
The display now reads 2.2546563 12. What kind of number is this? This is an example of a
whole number written in scientific notation. We'll study this concept when we get to decimal
numbers.
Practice Set E
Use a calculator to perform each multiplication.
Exercise:
Problem: 52 x 27
Solution:
1,404
Exercise:
Problem: 1,448 x 6,155
Solution:
8,912,440
Exercise:
Problem: 8,940,000 x 205,000
Solution:
1,832,700,000,000
Exercises
For the following problems, perform the multiplications. You may check each product with a
calculator.
Exercise:
Problem:
x3
Solution:
24
Exercise:
3
Problem:
x5
Exercise:
8
Problem:
x6
Solution:
48
Exercise:
5
Problem:
x7
Exercise:
Problem:6 x 1
Solution:
6
Exercise:
Problem:4 x 5
Exercise:
Problem:75 x 3
Solution:
225
Exercise:
Problem:35 x 5
Exercise:
45
Problem:
x 6
Solution:
270
Exercise:
31
Problem:
x 7
Exercise:
97
Problem:
x 6
Solution:
582
Exercise:
Probl 2
roblem:
x57
Exercise:
64
Problem:
x15
Solution:
960
Exercise:
73
Problem:
x15
Exercise:
81
Problem:
x95
Solution:
7,695
Exercise:
Problem:
x
Exercise:
Problem:57 x 64
Solution:
3,648
Exercise:
Problem:76 x 42
Exercise:
Problem:894 x 52
Solution:
46,488
Exercise:
Problem:684 x 38
Exercise:
115
Problem:
x. 22
Solution:
2,530
Exercise:
706
Problem:
x 8&1
Exercise:
328
Problem:
x
Solution:
6,888
Exercise:
Probl on
Tr m:
as x 94
Exercise:
Problem:930 x 26
Solution:
24,180
Exercise:
Problem:318 x 63
Exercise:
Beebe 582
ro em: 127
Solution:
73,914
Exercise:
24
Problem: f
x116
Exercise:
Problem: =P
x 225
Solution:
68,625
Exercise:
Probl 782
m:
OR Sapa
Exercise:
1
Problem: ie
x 663
Solution:
511,173
Exercise:
Problem: age
x516
Exercise:
Problem:1,905 x 710
Solution:
1,352,550
Exercise:
Problem:5,757 x 5,010
Exercise:
Se 3,106
Pere TSO
Solution:
5,441,712
Exercise:
bl 9,300
Problem: «1,130
Exercise:
bl 7,057
Problem: «5,229
Solution:
36,901,053
Exercise:
; 8,051
Problem: x 5,580
Exercise:
bl 5,804
Pro em: 4.300
Solution:
24,957,200
Exercise:
Problem:
roblem ale
Exercise:
724
Problem:
ro em: | 0
Solution:
0
Exercise:
problem: 2°049
Tr em:
7 ye AL
Exercise:
_ 5,173
Problem:
x 8
Solution:
41,384
Exercise:
Problem: nee
x 0
Exercise:
Problem: £2008
x 0
Solution:
0
Exercise:
1
Problem: le
x 142
Exercise:
Problem: ape
x 190
Solution:
73,530
Exercise:
Probl 3,400
Tr m:
oble 70
Exercise:
pean 460,000
roore™ 14,000
Solution:
6,440,000,000
Exercise:
, 558,000,000
Problem: | 81,000
Exercise:
37,000
Problem: :
Bese AG
Solution:
4,440,000
Exercise:
498,000
Problem:
x 0
Exercise:
rer 4,585,000
ro em: 140
Solution:
641,900,000
Exercise:
Beanie 30,700,000
ro em: 180
Exercise:
Probl 8,000
m:
ciara x 10
Solution:
80,000
Exercise:
Problem:
Suppose a theater holds 426 people. If the theater charges $4 per ticket and sells every seat,
how much money would they take in?
Exercise:
Problem:
In an English class, a student is expected to read 12 novels during the semester and prepare a
report on each one of them. If there are 32 students in the class, how many reports will be
prepared?
Solution:
384 reports
Exercise:
Problem:
In a mathematics class, a final exam consists of 65 problems. If this exam is given to 28
people, how many problems must the instructor grade?
Exercise:
Problem:
A business law instructor gives a 45 problem exam to two of her classes. If each class has 37
people in it, how many problems will the instructor have to grade?
Solution:
3,330 problems
Exercise:
Problem:
An algebra instructor gives an exam that consists of 43 problems to four of his classes. If the
classes have 25, 28, 31, and 35 students in them, how many problems will the instructor
have to grade?
Exercise:
Problem:
In statistics, the term "standard deviation" refers to a number that is calculated from certain
data. If the data indicate that one standard deviation is 38 units, how many units is three
standard deviations?
Solution:
114 units
Exercise:
Problem:
Soft drinks come in cases of 24 cans. If a supermarket sells 857 cases during one week, how
many individual cans were sold?
Exercise:
Problem:
There are 60 seconds in 1 minute and 60 minutes in 1 hour. How many seconds are there in
1 hour?
Solution:
3,600 seconds
Exercise:
Problem:
There are 60 seconds in 1 minute, 60 minutes in one hour, 24 hours in one day, and 365 days
in one year. How many seconds are there in 1 year?
Exercise:
Problem:
Light travels 186,000 miles in one second. How many miles does light travel in one year?
(Hint: Can you use the result of the previous problem?)
Solution:
5,865,696,000,000 miles per year
Exercise:
Problem:
An elementary school cafeteria sells 328 lunches every day. Each lunch costs $1. How much
money does the cafeteria bring in in 2 weeks?
Exercise:
Problem:
A computer company is selling stock for $23 a share. If 87 people each buy 55 shares, how
much money would be brought in?
Solution:
$110,055
Exercises for Review
Exercise:
Problem: ({link]) In the number 421,998, how may ten thousands are there?
Exercise:
Problem: (({link]) Round 448,062,187 to the nearest hundred thousand.
Solution:
448,100,000
Exercise:
Problem: (({link]) Find the sum. 22,451 + 18,976.
Exercise:
Problem: (({link]) Subtract 2,289 from 3,001.
Solution:
712
Exercise:
Problem:
([link]) Specify which property of addition justifies the fact that (a first whole number + a
second whole number) = (the second whole number + the first whole number)
Concepts of Division of Whole Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to divide whole numbers. By
the end of the module students should be able to understand the process of
division, understand division of a nonzero number into zero, understand
why division by zero is undefined, and use a calculator to divide one whole
number by another.
Section Overview
e Division
e Division into Zero (Zero As a Dividend: a a #0)
e Division by Zero (Zero As a Divisor: . a#0)
e Division by and into Zero (Zero As a Dividend and Divisor: )
e Calculators
Division
Division is a description of repeated subtraction.
In the process of division, the concern is how many times one number is
contained in another number. For example, we might be interested in how
many 5's are contained in 15. The word times is significant because it
implies a relationship between division and multiplication.
There are several notations used to indicate division. Suppose Q records the
number of times 5 is contained in 15. We can indicate this by writing
5)15 5
; 15 divided by 5
5 into 15
15/5=Q 15+5=Q
15 divided by 5 15 divided by 5
Each of these division notations describes the same number, represented
here by the symbol @. Each notation also converts to the same
multiplication form. It is 15 = 5 x Q
In division,
Dividend
the number being divided into is called the dividend.
Divisor
the number dividing into the dividend is the divisor.
Quotient
the result of the division is called the quotient.
quotient
divisor \dividend
dividend
divisor = quotient
dividend/divisor = quotient dividend + divisor = quotient
Sample Set A
Find the following quotients using multiplication facts.
Example:
18 = 6
Sivice: 6X3. — ES:
18 +6=3
Notice also that
—6 Repeated subtraction
Thus, 6 is contained in 18 three times.
Example:
24
3
Since 3 x 8 = 24,
pas
3} =
Notice also that 3 could be subtracted exactly 8 times from 24. This
implies that 3 is contained in 24 eight times.
Example:
36
6
Since 6 x 6 = 36,
36) =
6 = 6
Thus, there are 6 sixes in 36.
Example:
9)72
Since 9 x 8 = 72,
8
9)72
Thus, there are 8 nines in 72.
Practice Set A
Use multiplication facts to determine the following quotients.
Exercise:
Problem: 32 — 8
Solution:
4
Exercise:
Problem: 18 — 9
Solution:
2
Exercise:
Problem:
Solution:
D
Exercise:
Problem:
Solution:
6
Exercise:
Problem:
Solution:
4
Exercise:
Problem: 4)36
Solution:
9
Division into Zero (Zero as a Dividend: o yas 0)
Let's look at what happens when the dividend (the number being divided
into) is zero, and the divisor (the number doing the dividing) is any whole
number except zero. The question is
: : 5
What number, if any, 18 any nonzero whole number *
Let's represent this unknown quotient by Q. Then,
aapromao trom =
any nonzero whole number ~~
Converting this division problem to its corresponding multiplication
problem, we get
0 = Q x (any nonzero whole number)
From our knowledge of multiplication, we can understand that if the
product of two whole numbers is zero, then one or both of the whole
numbers must be zero. Since any nonzero whole number is certainly not
zero, @ must represent zero. Then,
= (0
any nonzero whole number
Zero Divided By Any Nonzero Whole Number Is Zero
Zero divided any nonzero whole number is zero.
Division by Zero (Zero as a Divisor: = , a # 0)
Now we ask,
any nonzero whole number 5
What number, if any, is 7
Letting Q represent a possible quotient, we get
any nonzero whole number __ Q
age
Converting to the corresponding multiplication form, we have
(any nonzero whole number) = Q x 0
Since Q x 0 = 0, (any nonzero whole number) = 0. But this is absurd.
This would mean that 6 = 0, or 37 = 0. A nonzero whole number cannot
equal 0! Thus,
any nonzero whole number
0 does not name a number
Division by Zero is Undefined
Division by zero does not name a number. It is, therefore, undefined.
Division by and Into Zero (Zero as a Dividend and Divisor: a)
We are now curious about zero divided by zero (2). If we let Q represent a
potential quotient, we get
0
9 7@
Converting to the multiplication form,
0=Qx0
This results in
0=—0
This is a statement that is true regardless of the number used in place of Q.
For example,
o = 5, since 0 = 5 x 0.
o = 31, since 0 = 31 x 0.
Se = 286, since 0 = 286 x 0.
A unique quotient cannot be determined.
Indeterminant
Since the result of the division is inconclusive, we say that - is
indeterminant.
- is Indeterminant
The division ~ is indeterminant.
Sample Set B
Perform, if possible, each division.
Example:
2. Since division by 0 does not name a whole number, no quotient exists,
and we state = is undefined
Example:
0)14. Since division by 0 does not name a defined number, no quotient
exists, and we state 0)14. is undefined
Example:
9)0. Since division into 0 by any nonzero whole number results in 0, we
i 0
ave 9)0
Example:
q. Since division into 0 by any nonzero whole number results in 0, we
have 2 ==a(i)
Practice Set B
Perform, if possible, the following divisions.
Exercise:
olor
Problem:
Solution:
undefined
Exercise:
=)
Problem:
Solution:
0
Exercise:
Problem: 0)0
Solution:
indeterminant
Exercise:
Problem: 0s
Solution:
undefined
Exercise:
Problem: a
Solution:
undefined
Exercise:
Problem: A.
Solution:
0
Calculators
Divisions can also be performed using a calculator.
Sample Set C
Example:
Divide 24 by 3.
Display Reads
Type 24 24
Press + 24
Type 3 3
Press = 8
The display now reads 8, and we conclude that 24 + 3 = 8.
Example:
Divide 0 by 7.
Display Reads
Type 0 0
Press = 0
Type 7 v7
Press = 0
The display now reads 0, and we conclude that 0 + 7 = 0.
Example:
Divide 7 by 0.
Since division by zero is undefined, the calculator should register some
kind of error message.
Display Reads
Type i 7
Press cr ih
Type 0 0
Press = Error
The error message indicates an undefined operation was attempted, in this
case, division by zero.
Practice Set C
Use a calculator to perform each division.
Exercise:
Problem: 35 — 7
Solution:
5
Exercise:
Problem: 56 ~ 8
Solution:
7
Exercise:
Problem: 0 — 6
Solution:
0
Exercise:
Problem: 3 — 0
Solution:
An error message tells us that this operation is undefined. The
particular message depends on the calculator.
Exercise:
Problem: 0 — 0
Solution:
An error message tells us that this operation cannot be performed.
Some calculators actually set 0 + 0 equal to 1. We know better! 0 + 0
is indeterminant.
Exercises
For the following problems, determine the quotients (if possible). You may
use a Calculator to check the result.
Exercise:
Problem: 4)32
Solution:
8
Exercise:
Problem: 7)42
Exercise:
Problem: 6)18
Solution:
3
Exercise:
Problem: 2)14
Exercise:
Problem:
Solution:
9
Exercise:
Problem:
Exercise:
Problem:
Solution:
7
Exercise:
Problem:
Exercise:
Problem:
Solution:
4
Exercise:
Problem:
Exercise:
Problem:
24+ 8
10+ 2
Solution:
5
Exercise:
Problem
Exercise:
Problem
:21+7
:21+3
Solution:
fi
Exercise:
Problem:
Exercise:
Problem:
Solution:
not defined
Exercise:
Problem:
Exercise:
Problem:
12+4
3)9
Solution:
3
Exercise:
Problem:
Exercise:
Problem:
Solution:
0
Exercise:
Problem:
Exercise:
Problem:
Solution:
fs)
Exercise:
Problem:
Exercise:
Problem:
Solution:
8
Exercise:
7)0
DO
co|o
Problem:
Exercise:
Problem: 72 — 8
Solution:
J
Exercise:
Problem: Write =~ = 8 using three different notations.
Exercise:
Problem: Write 22 = 3 using three different notations.
9
Solution:
27+9=3;9)07 =3; 2% =3
Exercise:
4
Problem: [n the statement 6)24
6 is called the .
24 is called the .
4 is called the .
Exercise:
Problem: In the statement 56 + 8 = 7,
7 is called the .
8 is called the .
56 is called the .
Solution:
7 is quotient; 8 is divisor; 56 is dividend
Exercises for Review
Exercise:
Problem: ({link]) What is the largest digit?
Exercise:
8,006
Problem: ([link]) Find the sum. +4118
Solution:
12,124
Exercise:
631
Problem: ({link]) Find the difference. 599
Exercise:
Problem:
({link]) Use the numbers 2, 3, and 7 to illustrate the associative
property of addition.
Solution:
(24+3)4+7=24(3+7)=12
5+7=2+10=12
Exercise:
Problem: (({link]) Find the product
86
12
Division of Whole Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to divide whole numbers. By
the end of the module students should be able to be able to divide a whole
number by a single or multiple digit divisor and interpret a calculator
statement that a division results in a remainder.
Section Overview
e Division with a Single Digit Divisor
e Division with a Multiple Digit Divisor
e Division with a Remainder
Calculators
Division with a Single Digit Divisor
Our experience with multiplication of whole numbers allows us to perform
such divisions as 75 + 5. We perform the division by performing the
corresponding multiplication, 5 x Q = 75. Each division we considered in
[link] had a one-digit quotient. Now we will consider divisions in which the
quotient may consist of two or more digits. For example, 75 + 5.
Let's examine the division 75 + 5. We are asked to determine how many 5's
are contained in 75. We'll approach the problem in the following way.
1. Make an educated guess based on experience with multiplication.
2. Find how close the estimate is by multiplying the estimate by 5.
3. If the product obtained in step 2 is less than 75, find out how much less
by subtracting it from 75.
4. If the product obtained in step 2 is greater than 75, decrease the
estimate until the product is less than 75. Decreasing the estimate
makes sense because we do not wish to exceed 75.
We can suggest from this discussion that the process of division consists of
The Four Steps in Division
1. an educated guess
2. a multiplication
3. a subtraction
4. bringing down the next digit (if necessary)
The educated guess can be made by determining how many times the
divisor is contained in the dividend by using only one or two digits of the
dividend.
Sample Set A
Example:
Find 75 + 5.
5)75 Rewrite the problem using a division bracket.
10
5)75
Make an educated guess by noting that one 5 is contained in 75 at most 10
times.
Since 7 is the tens digit, we estimate that 5 goes into 75 at most 10 times.
10
5)75
=50
25
Now determine how close the estimate is.
10 fives is 10 x 5 = 50. Subtract 50 from 75.
Estimate the number of 5's in 25.
There are exactly 5 fives in 25.
5 10 fives + 5 fives = 15 fives.
10
5)75
=o)
25
—25
0
There are 15 fives contained in 75.
Check:
75215X5
75 ~ 75
Thus, 75 +5 = 15.
The notation in this division can be shortened by writing.
15
5)75
ah
25
—25
0
Divide: 5 goes into 7 at most 1 time.
Multiply: 1 x 5=5. Write 5 below 7.
Subtract: 7-5 = 2. Bring down the 5.
Divide: 5 goes into 25 exactly 5 times.
Multiply: 5 x 5 = 25. Write 25 below 25.
Subtract? 25-25 =;
Example:
Find 4,944 = 8.
8 4944
Rewrite the problem using a division bracket.
600
8 )4944
—4800
144
8 goes into 49 at most 6 times, and 9 is in the hundreds column. We'll
guess 600.
Then, 8 x 600 = 4800.
10
600
8 \4944
—4800
144
— 80
64
8 goes into 14 at most 1 time, and 4 is in the tens column. We'll guess 10.
8 goes into 64 exactly 8 times.
600 eights + 10 eights + 8 eights = 618 eights.
Check:
494428 X618
4944 ~ 4944
Thus, 4,944 + 8 = 618.
As in the first problem, the notation in this division can be shortened by
eliminating the subtraction signs and the zeros in each educated guess.
Divide: 8 goes into 49 at most 6 times.
Multiply: 6 x 8 = 48. Write 48 below 49.
Subtract: 49 - 48 = 1. Bring down the 4.
Divide: 8 goes into 14 at most 1 time.
Multiply: 1 x 8 = 8. Write 8 below 14.
| Subtract: 14-8 —=6. Bring down the 4.
Divide: 8 goes into 64 exactly 8 times.
Multiply: 8 x 8 = 64. Write 64 below 64.
Subtract: 64-64=0.
Note: Not all divisions end in zero. We will examine such divisions in a
subsequent subsection.
Practice Set A
Perform the following divisions.
Exercise:
Problem: 126 ~ 7
Solution:
18
Exercise:
Problem: 324 — 4
Solution:
81
Exercise:
Problem: 2,559 =~ 3
Solution:
853
Exercise:
Problem: 5,645 =~ 5
Solution:
1,129
Exercise:
Problem: 757,125 ~ 9
Solution:
84,125
Division with a Multiple Digit Divisor
The process of division also works when the divisor consists of two or more
digits. We now make educated guesses using the first digit of the divisor
and one or two digits of the dividend.
Sample Set B
Example:
Find 2,232 + 36.
36) 2232
Use the first digit of the divisor and the first two digits of the dividend to
make the educated guess.
3 goes into 22 at most 7 times.
Try 7: 7 X 36 = 252 which is greater than 223. Reduce the estimate.
Try 6: 6 x 36 = 216 which is less than 223.
6
36) 2232
—216|
72
Multiply: 6 x 36 = 216. Write 216 below 223.
Subtract: 223 - 216 = 7. Bring down the 2.
Divide 3 into 7 to estimate the number of times 36 goes into 72. The 3 goes
into 7 at most 2 times.
pede OX OF = fs
62
36)2232
2164
72
-72
0
Check:
2232 2 36 X 62
2232 ~ 2232
Thus, 2,232 + 36 = 62.
Example:
Find 2,417,228 + 802.
802 )2417228
First, the educated guess: 24 + 8 = 3. Then 3 x 802 = 2406, which is
less than 2417. Use 3 as the guess. Since 3 x 802 = 2406, and 2406 has
four digits, place the 3 above the fourth digit of the dividend.
3
802) 2417228
— 2406]
112
Subtract: 2417 - 2406 = 11.
Bring down the 2.
The divisor 802 goes into 112 at most 0 times. Use 0.
30
802) 2417228
— 2406!
112
—0y
1122
Multiply: Oe 2802 — Or
Subtract: jee Orem 1 ILA
Bring down the 2.
The 8 goes into 11 at most 1 time, and 1 x 802 = 802, which is less than
i ry: A
Subtract 1122 — 802 = 320
Bring down the 8.
8 goes into 32 at most 4 times.
A x 802 = 3208.
Use 4.
3014
802 ) 2417228
Check:
2417228 2 3014 X 802
2417228 ~ 2417228
Thus, 2,417,228 + 802 = 3,014.
Practice Set B
Perform the following divisions.
Exercise:
Problem: 1,376 ~ 32
Solution:
43
Exercise:
Problem: 6,160 ~ 55
Solution:
112
Exercise:
Problem: 18,605 ~ 61
Solution:
305
Exercise:
Problem: 144,768 ~ 48
Solution:
3,016
Division with a Remainder
We might wonder how many times 4 is contained in 10. Repeated
subtraction yields
10
6
—4
Since the remainder is less than 4, we stop the subtraction. Thus, 4 goes into
10 two times with 2 remaining. We can write this as a division as follows.
2
4)10
— 8
2
Divide: A goes into 10 at most 2 times.
Multiply: 2 x 4=8. Write 8 below 0.
Subtract: 10-8 = 2.
Since 4 does not divide into 2 (the remainder is less than the divisor) and
there are no digits to bring down to continue the process, we are done. We
write
2R2
4) Morl+4= eRe
) 2 with remainder 2
Sample Set C
Example:
Find 85 + 3.
w
AIS Sle Bly
Divide: 3 goes into 8 at most 2 times.
Multiply: 2 x 3=6. Write 6 below 8.
Subtract: 8-6 = 2. Bring down the 5.
Divide: 3 goes into 25 at most 8 times.
Multiply: 3 x 8 = 24. Write 24 below 25.
Subtract: 25-24=1.
There are no more digits to bring down to continue the process. We are
done. One is the remainder.
Check: Multiply 28 and 3, then add 1.
28
x_ 3
84
rel
85
Thus, 89 + 3 = 28R1.
Example:
Find 726 + 23.
Check: Multiply 31 by 23, then add 13.
31
X 23
93
62_
713
+ 13
726
Thus, 726 + 23 = 31 R13.
Practice Set C
Perform the following divisions.
Exercise:
Problem:75 — 4
Solution:
18 R3
Exercise:
Problem:346 — 8
Solution:
43 R2
Exercise:
Problem:489 ~ 21
Solution:
23 R6
Exercise:
Problem:5,016 + 82
Solution:
61 R14
Exercise:
Problem:41,196 ~ 67
Solution:
614 R58
Calculators
The calculator can be useful for finding quotients with single and multiple
digit divisors. If, however, the division should result in a remainder, the
calculator is unable to provide us with the particular value of the remainder.
Also, some calculators (most nonscientific) are unable to perform divisions
in which one of the numbers has more than eight digits.
Sample Set D
Use a calculator to perform each division.
Example:
328 = 8
Type 328
Press =
Type 8
Press =
The display now reads 41.
Example:
53,136 + 82
Type 593136
Press ae
Type 82
Press a
The display now reads 648.
Example:
730,019,001 + 326
We first try to enter 730,019,001 but find that we can only enter 73001900.
If our calculator has only an eight-digit display (as most nonscientific
calculators do), we will be unable to use the calculator to perform this
division.
Example:
3727 + 49
Type 277
Press ice
Type 49
Press =
The display now reads 76.061224.
This number is an example of a decimal number (see [link]). When a
decimal number results in a calculator division, we can conclude that the
division produces a remainder.
Practice Set D
Use a calculator to perform each division.
Exercise:
Problem: 3,330 + 74
Solution:
45
Exercise:
Problem: 63,365 ~ 115
Solution:
dol
Exercise:
Problem: 21,996,385,287 + 53
Solution:
Since the dividend has more than eight digits, this division cannot be
performed on most nonscientific calculators. On others, the answer is
415,026,137.4
Exercise:
Problem: 4,558 = 67
Solution:
This division results in 68.02985075, a decimal number, and therefore,
we cannot, at this time, find the value of the remainder. Later, we will
discuss decimal numbers.
Exercises
For the following problems, perform the divisions.
The first 38 problems can be checked with a calculator by multiplying the
divisor and quotient then adding the remainder.
Exercise:
Problem: 52 — 4
Solution:
13
Exercise:
Problem: 776 — 8
Exercise:
Problem: 603 ~ 9
Solution:
67
Exercise:
Problem
Exercise:
Problem
: 240 = 8
: 208 + 4
Solution:
52
Exercise:
Problem
Exercise:
Problem
576 + 6
:21+7
Solution:
3
Exercise:
Problem:
Exercise:
Problem
: 140 = 2
Solution:
70
Exercise:
Problem:
Exercise:
Problem:
Solution:
61
Exercise:
Problem:
Exercise:
Problem:
Solution:
og
Exercise:
Problem:
Exercise:
Problem:
Solution:
67
Exercise:
Problem:
528 + 8
244 + 4
0+7
Lips
96=8
orl
896 + 56
Exercise:
Problem
Solution:
87
Exercise:
Problem
Exercise:
Problem
Solution:
04
Exercise:
Problem
Exercise:
Problem
Solution:
a2
Exercise:
Problem
Exercise:
1,044 + 12
988 + 19
: 5,238 + 97
£2,530 + 55
: 4,264 + 82
: 637 + 13
Problem:
Solution:
38
Exercise:
Problem:
Exercise:
Problem:
Solution:
45
Exercise:
Problem:
Exercise:
Problem:
Solution:
777
Exercise:
Problem:
Exercise:
Problem:
3,420 + 90
5,655 + 87
F115 47
9,328 + 22
55,167 + 71
68,356 + 92
27702 = 81
Solution:
342
Exercise:
Problem: 6,510 + 31
Exercise:
Problem: 60,536 + 94
Solution:
644
Exercise:
Problem: 31,844 ~ 38
Exercise:
Problem: 23,985 ~ 45
Solution:
see,
Exercise:
Problem: 60,606 ~ 74
Exercise:
Problem: 2,975,400 + 285
Solution:
10,440
Exercise:
Problem: 1,389,660 ~ 795
Exercise:
Problem: 7,162,060 = 879
Solution:
8,147 remainder 847
Exercise:
Problem: 7,561,060 ~ 909
Exercise:
Problem: 38 —~ 9
Solution:
4 remainder 2
Exercise:
Problem: 97 — 4
Exercise:
Problem: 199 ~ 3
Solution:
66 remainder 1
Exercise:
Problem: 573 — 6
Exercise:
Problem: 10,701 ~ 13
Solution:
823 remainder 2
Exercise:
Problem: 13,521 + 53
Exercise:
Problem: 3,628 ~ 90
Solution:
4O remainder 28
Exercise:
Problem: 10,592 ~ 43
Exercise:
Problem: 19,965 ~ 30
Solution:
665 remainder 15
Exercise:
Problem: 8,320 + 21
Exercise:
Problem: 61,282 ~ 64
Solution:
957 remainder 34
Exercise:
Problem: 1,030 + 28
Exercise:
Problem: 7,319 + 11
Solution:
665 remainder 4
Exercise:
Problem: 3,628 ~ 90
Exercise:
Problem: 35,279 ~ 77
Solution:
458 remainder 13
Exercise:
Problem: 52,196 + 55
Exercise:
Problem: 67,751 ~ 68
Solution:
996 remainder 23
For the following 5 problems, use a calculator to find the quotients.
Exercise:
Problem: 4,346 ~ 53
Exercise:
Problem: 3,234 + 77
Solution:
42
Exercise:
Problem: 6,771 ~ 37
Exercise:
Problem: 4,272,320 + 520
Solution:
8,216
Exercise:
Problem: 7,558,110 + 651
Exercise:
Problem:
A mathematics instructor at a high school is paid $17,775 for 9
months. How much money does this instructor make each month?
Solution:
$1,975 per month
Exercise:
Problem:
A couple pays $4,380 a year for a one-bedroom apartment. How much
does this couple pay each month for this apartment?
Exercise:
Problem:
Thirty-six people invest a total of $17,460 in a particular stock. If they
each invested the same amount, how much did each person invest?
Solution:
$485 each person invested
Exercise:
Problem:
Each of the 28 students in a mathematics class buys a textbook. If the
bookstore sells $644 worth of books, what is the price of each book?
Exercise:
Problem:
A certain brand of refrigerator has an automatic ice cube maker that
makes 336 ice cubes in one day. If the ice machine makes ice cubes at
a constant rate, how many ice cubes does it make each hour?
Solution:
14 cubes per hour
Exercise:
Problem:
A beer manufacturer bottles 52,380 ounces of beer each hour. If each
bottle contains the same number of ounces of beer, and the
manufacturer fills 4,365 bottles per hour, how many ounces of beer
does each bottle contain?
Exercise:
Problem:
A computer program consists of 68,112 bits. 68,112 bits equals 8,514
bytes. How many bits in one byte?
Solution:
8 bits in each byte
Exercise:
Problem:
A 26-story building in San Francisco has a total of 416 offices. If each
floor has the same number of offices, how many floors does this
building have?
Exercise:
Problem:
A college has 67 classrooms and a total of 2,546 desks. How many
desks are in each classroom if each classroom has the same number of
desks?
Solution:
38
Exercises for Review
Exercise:
Problem: ({link]) What is the value of 4 in the number 124,621?
Exercise:
Problem: ({link]) Round 604,092 to the nearest hundred thousand.
Solution:
600,000
Exercise:
Problem: ({link]) What whole number is the additive identity?
Exercise:
Problem: ({link]) Find the product. 6,256 x 100.
Solution:
625,600
Exercise:
Problem: ({link]) Find the quotient. 0 + 11.
Some Interesting Facts about Division
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses interesting facts about diving
whole numbers. By the end of the module students should be able to
recognize a whole number that is divisible by 2, 3, 4, 5, 6, 8, 9, or 10.
Section Overview
e Division by 2, 3, 4, and 5
e Division by 6, 8, 9, and 10
Quite often, we are able to determine if a whole number is divisible by
another whole number just by observing some simple facts about the
number. Some of these facts are listed in this section.
Division by 2, 3, 4, and 5
Division by 2
A whole number is divisible by 2 if its last digit is 0, 2, 4, 6, or 8.
The numbers 80, 112, 64, 326, and 1,008 are all divisible by 2 since the last
digit of each is 0, 2, 4, 6, or 8, respectively.
The numbers 85 and 731 are not divisible by 2.
Division by 3
A whole number is divisible by 3 if the sum of its digits is divisible by 3.
The number 432 is divisible by 3 since 4 + 3 + 2 = 9 and 9 is divisible by
2
432 +3 = 144
The number 25 is not divisible by 3 since 2 + 5 = 7, and 7 is not divisible
by 3:
Division by 4
A whole number is divisible by 4 if its last two digits form a number that is
divisible by 4.
The number 31,048 is divisible by 4 since the last two digits, 4 and 8, form
a number, 48, that is divisible by 4.
31048 + 4 = 7262
The number 137 is not divisible by 4 since 37 is not divisible by 4.
Division by 5
A whole number is divisible by 5 if its /ast digit is 0 or 5.
Sample Set A
Example:
The numbers 65, 110, 8,030, and 16,955 are each divisible by 5 since the
last digit of each is 0 or 5.
Practice Set A
State which of the following whole numbers are divisible by 2, 3, 4, or 5. A
number may be divisible by more than one number.
Exercise:
Problem: 26
Solution:
2
Exercise:
Problem: 81
Solution:
3
Exercise:
Problem: 51
Solution:
3
Exercise:
Problem: 385
Solution:
5
Exercise:
Problem: 6,112
Solution:
Ded
Exercise:
Problem: 470
Solution:
25D
Exercise:
Problem: 113,154
Solution:
2,0
Division by 6, 8, 9, 10
Division by 6
A number is divisible by 6 if it is divisible by both 2 and 3.
The number 234 is divisible by 2 since its last digit is 4. It is also divisible
by 3 since 2+ 3+ 4 = 9 and 9 is divisible by 3. Therefore, 234 is divisible
by 6.
The number 6,532 is not divisible by 6. Although its last digit is 2, making
it divisible by 2, the sum of its digits,6 + 5+ 3+ 2 = 16, and 16 is not
divisible by 3.
Division by 8
A whole number is divisible by 8 if its last three digits form a number that
is divisible by 8.
The number 4,000 is divisible by 8 since 000 is divisible by 8.
The number 13,128 is divisible by 8 since 128 is divisible by 8.
The number 1,170 is not divisible by 8 since 170 is not divisible by 8.
Division by 9
A whole number is divisible by 9 if the sum of its digits is divisible by 9.
The number 702 is divisible by 9 since 7 + 0 + 2 is divisible by 9.
The number 6588 is divisible by 9 since 6 + 5 + 8 + 8 = 27 is divisible by
9.
The number 14,123 is not divisible by 9 since 1+4+1+2+3=11is
not divisible by 9.
Division by 10
A Whole number is divisible by 10 if its last digit is 0.
Sample Set B
Example:
The numbers 30, 170, 16,240, and 865,000 are all divisible by 10.
Practice Set B
State which of the following whole numbers are divisible 6, 8, 9, or 10.
Some numbers may be divisible by more than one number.
Exercise:
Problem: 900
Solution:
6, 9, 10
Exercise:
Problem: 6,402
Solution:
6
Exercise:
Problem: 6,660
Solution:
6, 9, 10
Exercise:
Problem: 55,116
Solution:
G39
Exercises
For the following 30 problems, specify if the whole number is divisible by
2, 3, 4, 5, 6, 8, 9, or 10. Write "none" if the number is not divisible by any
digit other than 1. Some numbers may be divisible by more than one
number.
Exercise:
Problem: 48
Solution:
23:4. 6.8
Exercise:
Problem: 85
Exercise:
Problem: 30
Solution:
2365). 0..10)
Exercise:
Problem:
Exercise:
Problem:
Solution:
2
Exercise:
Problem:
Exercise:
Problem:
Solution:
2/4
Exercise:
Problem:
Exercise:
Problem:
Solution:
3
Exercise:
Problem:
83
98
972
892
676
903
800
Exercise:
Problem: 223
Solution:
none
Exercise:
Problem: 836
Exercise:
Problem: 665
Solution:
5
Exercise:
Problem: 4,381
Exercise:
Problem: 2,195
Solution:
5
Exercise:
Problem: 2,544
Exercise:
Problem: 5,172
Solution:
2,3, 4,6
Exercise:
Problem: 1,307
Exercise:
Problem: 1,050
Solution:
Zoos Uy LO
Exercise:
Problem: 3,898
Exercise:
Problem: 1,621
Solution:
none
Exercise:
Problem: 27,808
Exercise:
Problem: 45,764
Solution:
2,4
Exercise:
Problem: 49,198
Exercise:
Problem: 296,122
Solution:
2
Exercise:
Problem: 178,656
Exercise:
Problem: 5,102,417
Solution:
none
Exercise:
Problem: 16,990,792
Exercise:
Problem: 620,157,659
Solution:
none
Exercise:
Problem: 457,687,705
Exercises for Review
Exercise:
Problem: ({link]) In the number 412, how many tens are there?
Solution:
1
Exercise:
Problem: ({link]) Subtract 613 from 810.
Exercise:
Problem: ([link]) Add 35, 16, and 7 in two different ways.
Solution:
(35 +16) +7=514+7=58
35 + (16 +7) = 35 +23 =58
Exercise:
Problem: ({link]) Find the quotient 35 + 0, if it exists.
Exercise:
Problem: ({link]) Find the quotient. 3654 + 42.
Solution:
87
Properties of Multiplication
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses
properties of multiplication of whole numbers. By the end of the module students should be able to understand and
appreciate the commutative and associative properties of multiplication and understand why 1 is the multiplicative
identity.
Section Overview
e The Commutative Property of Multiplication
e The Associative Property of Multiplication
e The Multiplicative Identity
We will now examine three simple but very important properties of multiplication.
The Commutative Property of Multiplication
Commutative Property of Multiplication
The product of two whole numbers is the same regardless of the order of the factors.
Sample Set A
Example:
Multiply the two whole numbers.
6
hal
6-7 = 42
7:6= 42
The numbers 6 and 7 can be multiplied in any order. Regardless of the order they are multiplied, the product is 42.
Practice Set A
Use the commutative property of multiplication to find the products in two ways.
Exercise:
Problem:
Solution:
15-6 = 90 and 6-15 = 90
Exercise:
Problem:
432
428
Solution:
432 - 428 = 184,896 and 428 - 432 = 184,896
The Associative Property of Multiplication
Associative Property of Multiplication
If three whole numbers are multiplied, the product will be the same if the first two are multiplied first and then that
product is multiplied by the third, or if the second two are multiplied first and that product is multiplied by the
first. Note that the order of the factors is maintained.
It is a common mathematical practice to use parentheses to show which pair of numbers is to be combined first.
Sample Set B
Example:
Multiply the whole numbers.
8
3
14
(8-3)-14= 24-14 = 336
8- (3-14) = 8-42 = 336
Practice Set B
Use the associative property of multiplication to find the products in two ways.
Exercise:
Problem:
Solution:
168
Exercise:
Problem:
73
18
126
Solution:
165,564
The Multiplicative Identity
The Multiplicative Identity is 1
The whole number 1 is called the multiplicative identity, since any whole number multiplied by 1 is not changed.
Sample Set C
Example:
Multiply the whole numbers.
12
1
Who I= IH
Iho WA = 1
Practice Set C
Multiply the whole numbers.
Exercise:
Problem:
843
Solution:
843
Exercises
For the following problems, multiply the numbers.
Exercise:
Problem:
Solution:
234
Exercise:
Problem:
18
41
Exercise:
Problem:
Solution:
4,032
Exercise:
Problem:
132
Exercise:
Problem:
1000
326
Solution:
326,000
Exercise:
Problem:
1400
Exercise:
Problem:
Solution:
252
Exercise:
Problem:
40
16
Exercise:
Problem:
Solution:
21,340
Exercise:
Problem:
110
85
0
Exercise:
Problem:
462
18
Solution:
8,316
Exercise:
Problem:
101
For the following 4 problems, show that the quantities yield the same products by performing the multiplications.
Exercise:
Problem: (4 - 8) - 2 and 4 - (8 - 2)
Solution:
32-2=—64=4-16
Exercise:
Problem: (100 - 62) - 4 and 100 - (62 - 4)
Exercise:
Problem: 23 - (11 - 106) and (23 - 11) - 106
Solution:
23 - 1,166 = 26,818 = 253 - 106
Exercise:
Problem: 1 - (5-2) and (1-5) - 2
Exercise:
Problem:
The fact that
(a first number - a second number) - a third number = a first number - (a second number - a third num
is an example of the property of multiplication.
Solution:
associative
Exercise:
Problem:
The fact that 1 - any number = that particular numberis an example of the property of multiplication.
Exercise:
Problem: Use the numbers 7 and 9 to illustrate the commutative property of multiplication.
Solution:
7-9=63=9-7
Exercise:
Problem: Use the numbers 6, 4, and 7 to illustrate the associative property of multiplication.
Exercises for Review
Exercise:
Problem: ({link]) In the number 84,526,098,441, how many millions are there?
Solution:
6
Exercise:
85
Problem: ({link]) Replace the letter m with the whole number that makes the addition true. + m™
97
Exercise:
Problem: ((link]) Use the numbers 4 and 15 to illustrate the commutative property of addition.
Solution:
4+15=19
15+4=19
Exercise:
Problem: ((link]) Find the product. 8,000,000 x 1,000.
Exercise:
Problem: ((link]) Specify which of the digits 2, 3, 4, 5, 6, 8,10 are divisors of the number 2,244.
Solution:
2,3,4,6
Summary of Key Concepts
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module summarizes the concepts discussed in the
chapter "Multiplication and Division of Whole Numbers."
Summary of Key Concepts
Multiplication ({link])
Multiplication is a description of repeated addition.
(es ont Mes ht eee
ad
7 appears 4 times
This expression is described by writing 4 X 7.
Multiplicand/Multiplier/Product ({link])
In a multiplication of whole numbers, the repeated addend is called the
multiplicand, and the number that records the number of times the
multiplicand is used is the multiplier. The result of the multiplication is the
product.
Factors ({link])
In a multiplication, the numbers being multiplied are also called factors.
Thus, the multiplicand and the multiplier can be called factors.
Division ((link])
Division is a description of repeated subtraction.
Dividend/Divisor/Quotient ({link])
In a division, the number divided into is called the dividend, and the
number dividing into the dividend is called the divisor. The result of the
division is called the quotient.
quotient
divisor ) dividend
Division into Zero ([link])
Zero divided by any nonzero whole number is zero.
Division by Zero ([link])
Division by zero does not name a whole number. It is, therefore, undefined.
The quotient 4 is indeterminant.
Division by 2, 3, 4, 5, 6, 8, 9, 10 ([link])
Division by the whole numbers 2, 3, 4, 5, 6, 8, 9, and 10 can be determined
by noting some certain properties of the particular whole number.
Commutative Property of Multiplication ({link])
The product of two whole numbers is the same regardless of the order of
the factors. 3x 5=5 x3
Associative Property of Multiplication ({link])
If three whole numbers are to be multiplied, the product will be the same if
the first two are multiplied first and then that product is multiplied by the
third, or if the second two are multiplied first and then that product is
multiplied by the first.
(3x5) x2=3 x (5 2}
Note that the order of the factors is maintained.
Multiplicative Identity ([link])
The whole number 1 is called the multiplicative identity since any whole
number multiplied by 1 is not changed.
4x1=4
1x4=4
Exercise Supplement
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is an exercise supplement for the chapter
"Multiplication and Division of Whole Numbers" and contains many
exercise problems. Odd problems are accompanied by solutions.
Exercise Supplement
Multiplication of Whole Numbers ((link])
Exercise:
Problem:
In the multiplication 5 x 9 = 45, 5 and 9 are called and 45 is called
the .
Solution:
factors; product
Exercise:
Problem:
In the multiplication 4 x 8 = 32, 4 and 8 are called and 32 is called
the .
Concepts of Division of Whole Numbers ([link])
Exercise:
Problem:
In the division 24 ~ 6 = 4, 6 is called the , and 4 is called the .
Solution:
divisor; quotient
Exercise:
Problem:
In the division 36 ~ 2 = 18, 2 is called the , and 18 is called the .
Some Interesting Facts about Division ([link])
Exercise:
Problem: A number is divisible by 2 only if its last digit is .
Solution:
an even digit (0, 2, 4, 6, or 8)
Exercise:
Problem:
A number is divisible by 3 only if of its digits is divisible by 3.
Exercise:
Problem:
A number is divisible by 4 only if the rightmost two digits form a
number that is .
Solution:
divisible by 4
Multiplication and Division of Whole Numbers ((link],[link])
Find each product or quotient.
Exercise:
24
Problem:
Exercise:
x3
14
Problem:
x 8
Solution:
112
Exercise:
Problem
Exercise:
Problem
:21+7
230+5
Solution:
7
Exercise:
36
Problem:
Exercise:
| DD
87
Problem:
" x35
Solution:
3,045
Exercise:
117
Problem:
x42
Exercise:
Problem: 208 ~ 52
Solution:
4
Exercise:
521
Problem:
Exercise:
Problem:
Solution:
15,075
Exercise:
Problem: 1338 ~ 446
Exercise:
Problem: 2814 — 201
Solution:
14
Exercise:
Problem:
Exercise:
5821
x8
6016
Problem:
ya |
Solution:
42,112
Exercise:
Problem
Exercise:
Problem
> 576 + 24
: 3969 + 63
Solution:
63
Exercise:
5482
Problem:
Exercise:
x $22
9104
Problem:
x 115
Solution:
1,046,960
Exercise:
6102
Problem:
x 1000
Exercise:
10101
Problem:
x 10000
Solution:
101,010,000
Exercise:
Problem: 162,006 ~ 31
Exercise:
Problem: 0 — 25
Solution:
0
Exercise:
Problem: 25 — 0
Exercise:
Problem: 4280 ~ 10
Solution:
428
Exercise:
Problem: 2126000 — 100
Exercise:
Problem: 84 — 15
Solution:
5 remainder 9
Exercise:
Problem: 126 ~ 4
Exercise:
Problem: 424 — 0
Solution:
not defined
Exercise:
Problem: 1198 ~ 46
Exercise:
Problem: 995 =~ 31
Solution:
32 remainder 3
Exercise:
Problem: 0 — 18
Exercise:
2162
Problem:
x 1421
Solution:
3,072,202
Exercise:
Problem: 0 x 0
Exercise:
Problem: 5 x 0
Solution:
0
Exercise:
Problem: 64 x 1
Exercise:
Problem: 1 x 0
Solution:
0
Exercise:
Problem: 0 — 3
Exercise:
Problem: 14 — 0
Solution:
not defined
Exercise:
Problem: 35 — 1
Exercise:
Problem: 1 — 1
Solution:
1
Properties of Multiplication ({link])
Exercise:
Problem:
Use the commutative property of multiplication to rewrite 36 x 128.
Exercise:
Problem:
Use the commutative property of multiplication to rewrite 114 x 226.
Solution:
226-114
Exercise:
Problem:
Use the associative property of multiplication to rewrite (5 - 4) - 8.
Exercise:
Problem:
Use the associative property of multiplication to rewrite 16 - (14 - 0).
Solution:
(16-14) -0
Multiplication and Division of Whole Numbers ((link],[link])
Exercise:
Problem:
A computer store is selling diskettes for $4 each. At this price, how
much would 15 diskettes cost?
Exercise:
Problem:
Light travels 186,000 miles in one second. How far does light travel in
23 seconds?
Solution:
4,278,000
Exercise:
Problem:
A dinner bill for eight people comes to exactly $112. How much
should each person pay if they all agree to split the bill equally?
Exercise:
Problem:
Each of the 33 students in a math class buys a textbook. If the
bookstore sells $1089 worth of books, what is the price of each book?
Solution:
$33
Proficiency Exam
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is a proficiency exam to the chapter
"Multiplication and Division of Whole Numbers." Each problem is
accompanied with a reference link pointing back to the module that
discusses the type of problem demonstrated in the question. The problems
in this exam are accompanied by solutions.
Proficiency Exam
Exercise:
Problem:
({link]) In the multiplication of 8 x 7 = 56, what are the names given
to the 8 and 7 and the 56?
Solution:
8 and 7 are factors; 56 is the product
Exercise:
Problem:
({link]) Multiplication is a description of what repeated process?
Solution:
Addition
Exercise:
Problem:
({link]) In the division 12 + 3 = 4, what are the names given to the 3
and the 4?
Solution:
3 is the divisor; 4 is the quotient
Exercise:
Problem:
({link]) Name the digits that a number must end in to be divisible by 2.
Solution:
0, 2, 4, 6, or 8
Exercise:
Problem:
({link]) Name the property of multiplication that states that the order of
the factors in a multiplication can be changed without changing the
product.
Solution:
commutative
Exercise:
Problem: ({link]) Which number is called the multiplicative identity?
Solution:
1
For problems 7-17, find the product or quotient.
Exercise:
Problem: ({link]) 14 x 6
Solution:
84
Exercise:
Problem: ({link]) 37 x 0
Solution:
0
Exercise:
Problem: ({link]) 352 x 1000
Solution:
352,000
Exercise:
Problem: (({link]) 5986 x 70
Solution:
419,020
Exercise:
Problem: ([link]) 12 x 12
Solution:
252
Exercise:
Problem: ((link]) 856 + 0
Solution:
not defined
Exercise:
Problem: ({link]) 0 + 8
Solution:
0
Exercise:
Problem: ({link]) 136 + 8
Solution:
17
Exercise:
Problem: ([link]) 432 + 24
Solution:
18
Exercise:
Problem: ([link]) 5286 + 37
Solution:
142 remainder 32
Exercise:
Problem: (({link]) 211 x 1
Solution:
211
For problems 18-20, use the numbers 216, 1,005, and 640.
Exercise:
Problem: ({link]) Which numbers are divisible by 3?
Solution:
216; 1,005
Exercise:
Problem: ({link]) Which number is divisible by 4?
Solution:
216; 640
Exercise:
Problem: ({link]) Which number(s) is divisible by 5?
Solution:
1,005; 640
Objectives
This module contains the learning objectives for the chapter "Exponents,
Roots, and Factorizations of Whole Numbers" from Fundamentals of
Mathematics by Denny Burzynski and Wade Ellis, jr.
After completing this chapter, you should
Exponents and Roots ({link])
e understand and be able to read exponential notation
e understand the concept of root and be able to read root notation
e be able to use a calculator having the y* key to determine a root
Grouping Symbols and the Order of Operations ({link])
e understand the use of grouping symbols
e understand and be able to use the order of operations
e use the calculator to determine the value of a numerical expression
Prime Factorization of Natural Numbers ({link])
e be able to determine the factors of a whole number
e be able to distinguish between prime and composite numbers
e be familiar with the fundamental principle of arithmetic
e be able to find the prime factorization of a whole number
The Greatest Common Factor (({link])
e be able to find the greatest common factor of two or more whole
numbers
The Least Common Multiple ({link])
e be able to find the least common multiple of two or more whole
numbers
Exponents and Roots
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses exponents and roots. By the end
of the module students should be able to understand and be able to read
exponential notation, understand the concept of root and be able to read root
notation, and use a calculator having the y* key to determine a root.
Section Overview
e Exponential Notation
e Reading Exponential Notation
e Roots
e Reading Root Notation
e Calculators
Exponential Notation
Exponential Notation
We have noted that multiplication is a description of repeated addition.
Exponential notation is a description of repeated multiplication.
Suppose we have the repeated multiplication
8-8-8-8-8
Exponent
The factor 8 is repeated 5 times. Exponential notation uses a superscript for
the number of times the factor is repeated. The superscript is placed on the
repeated factor, 8°, in this case. The superscript is called an exponent.
The Function of an Exponent
An exponent records the number of identical factors that are repeated in a
multiplication.
Sample Set A
Write the following multiplication using exponents.
Example:
3-3. Since the factor 3 appears 2 times, we record this as
32
Example:
62 - 62-62-62 - 62-62-62 - 62 - 62. Since the factor 62 appears 9 times,
we record this as
62°
Expand (write without exponents) each number.
Example:
12, ihe exponent 4 is recording 4 factors of 12 in a multiplication. Thus,
Telesis PS
Example:
706°. The exponent 3 is recording 3 factors of 706 in a multiplication.
Thus,
706° = 706 - 706 - 706
Practice Set A
Write the following using exponents.
Exercise:
Problem: 37 - 37
Solution:
a7?
Exercise:
Problem: 16-16-16-16-16
Solution:
16°
Exercise:
Problem: 9-9-9-9-9-9-9-9-9-9
Solution:
910
Write each number without exponents.
Exercise:
Problem: 85°
Solution:
85-85-85
Exercise:
Problem: 4’
Solution:
4-4-4-4-4.-4.4
Exercise:
Problem: 1,739”
Solution:
1,739 - 1,739
Reading Exponential Notation
In a number such as 8°,
Base
8 is called the base.
Exponent, Power
5 is called the exponent, or power. 8° is read as "eight to the fifth power,"
or more simply as "eight to the fifth," or "the fifth power of eight."
Squared
When a whole number is raised to the second power, it is said to be
squared. The number 5? can be read as
5 to the second power, or
5 to the second, or
5 squared.
Cubed
When a whole number is raised to the third power, it is said to be cubed.
The number 5? can be read as
5 to the third power, or
5 to the third, or
5 cubed.
When a whole number is raised to the power of 4 or higher, we simply say
that that number is raised to that particular power. The number 58 can be
read as
5 to the eighth power, or just
5 to the eighth.
Roots
In the English language, the word "root" can mean a source of something.
In mathematical terms, the word "root" is used to indicate that one number
is the source of another number through repeated multiplication.
Square Root
We know that 49 = 7”, that is, 49 = 7 - 7. Through repeated multiplication,
7 is the source of 49. Thus, 7 is a root of 49. Since two 7's must be
multiplied together to produce 49, the 7 is called the second or square root
of 49.
Cube Root
We know that 8 = 2°, that is, 8 = 2-2-2. Through repeated
multiplication, 2 is the source of 8. Thus, 2 is a root of 8. Since three 2's
must be multiplied together to produce 8, 2 is called the third or cube root
of 8.
We can continue this way to see such roots as fourth roots, fifth roots, sixth
roots, and so on.
Reading Root Notation
There is a symbol used to indicate roots of a number. It is called the radical
sign an
The Radical Sign of
The symbol is called a radical sign and indicates the nth root of a
number.
We discuss particular roots using the radical sign as follows:
Square Root
number indicates the square root of the number under the radical sign.
It is customary to drop the 2 in the radical sign when discussing square
roots. The symbol ,/— is understood to be the square root radical sign.
/49 =7 since 7-7 = 72 = 49
Cube Root
number indicates the cube root of the number under the radical sign.
4/8 = 2since2-2-2—223 —8
Fourth Root
Vv number indicates the fourth root of the number under the radical sign.
V7 81 = 3since3-3-3-3=34 = 81
In an expression such as 1/32
Radical Sign
/__ is called the radical sign.
Index
5 is called the index. (The index describes the indicated root.)
Radicand
32 is called the radicand.
Radical
¥ 32 is called a radical (or radical expression).
Sample Set B
Find each root.
Example:
/25 To determine the square root of 25, we ask, "What whole number
squared equals 25?" From our experience with multiplication, we know
this number to be 5. Thus,
/25=5
Check: 5-5 = 5% = 25
Example:
4/32 To determine the fifth root of 32, we ask, "What whole number raised
to the fifth power equals 32°?" This number is 2.
4/32 = 2
CHECK 27 e220) ere oP
Practice Set B
Find the following roots using only a knowledge of multiplication.
Exercise:
Problem: / 64
Solution:
8
Exercise:
Problem: / 100
Solution:
10
Exercise:
Problem: </ 64
Solution:
4
Exercise:
Problem: </ 64
Solution:
2
Calculators
Calculators with the ,/z, y*, and 1/z keys can be used to find or
approximate roots.
Sample Set C
Example: —_
Use the calculator to find A {aah
Display Reads
Type 121 121
Press fx 11
Example:
Find V/2187.
Display Reads
Type 2187 2187
Press y” 2187
Type i i
Press Le 14285714
Press = 3
¥/ 2187 = 3 (Which means that 3’ = 2187 .)
Practice Set C
Use a calculator to find the following roots.
Exercise:
Problem: </ 729
Solution:
9
Exercise:
Problem: */8503056
Solution:
54
Exercise:
Problem: \/53361
Solution:
Zak
Exercise:
Problem: \/16777216
Solution:
4
Exercises
For the following problems, write the expressions using exponential
notation.
Exercise:
Problem: 4 - 4
Solution:
42
Exercise:
Problem: 12 - 12
Exercise:
Problem: 9-9-9-9
Solution:
o4
Exercise:
Problem: 10-10-10-10-10-10
Exercise:
Problem: 826 - 826 - 826
Solution:
826°
Exercise:
Problem: 3,021 - 3,021 - 3,021 - 3,021 - 3,021
Exercise:
Problem: 6-6-:::: 6
85 factors of 6
Solution:
685
Exercise:
Problem: i 2
112 factors of 2
Exercise:
Problem: 1-1---- il
3,008 factors of 1
Solution:
1 3008
For the following problems, expand the terms. (Do not find the actual
value.)
Exercise:
Problem: 52
Exercise:
Problem: 7*
Solution:
Clete
Exercise:
Problem: 157
Exercise:
Problem: 117°
Solution:
117-117-117-117-117
Exercise:
Problem: 61°
Exercise:
Problem: 30°
Solution:
30 - 30
For the following problems, determine the value of each of the powers. Use
a calculator to check each result.
Exercise:
Problem: 32
Exercise:
Problem: 42
Solution:
4-4=16
Exercise:
Problem: 12
Exercise:
Problem: 107
Solution:
10-10 = 100
Exercise:
Problem: 117
Exercise:
Problem: 197
Solution:
12-12 = 144
Exercise:
Problem: 137
Exercise:
Problem: 157
Solution:
15-15 = 225
Exercise:
Problem: 14
Exercise:
Problem: 34
Solution:
3°3-3-3=—81
Exercise:
Problem: 7°
Exercise:
Problem: 10°
Solution:
10-10-10 = 1,000
Exercise:
Problem: 1007
Exercise:
Problem: 8°
Solution:
8-8-8 =—512
Exercise:
Problem: 5°
Exercise:
Problem: 92
Solution:
9-9-9 = 729
Exercise:
Problem: 62
Exercise:
Problem: 7!
Solution:
7i=7
Exercise:
Problem: 12°
Exercise:
Problem: 2°
Solution:
22024 222° 22 = 128
Exercise:
Problem: 0°
Exercise:
Problem: 84
Solution:
8-8-8-8 = 4,096
Exercise:
Problem: 5°
Exercise:
Problem: 6°
Solution:
6:6:-6-6-6-6-6-6-6 = 10,077,696
Exercise:
Problem: 25°
Exercise:
Problem: 427
Solution:
42-42 = 1,764
Exercise:
Problem: 31°
Exercise:
Problem: 15°
Solution:
15-15-15-15-15 = 759,375
Exercise:
Problem: 22°
Exercise:
Problem: 8167
Solution:
816 - 816 = 665,856
For the following problems, find the roots (using your knowledge of
multiplication). Use a calculator to check each result.
Exercise:
Problem: J 9
Exercise:
Problem: / 16
Solution:
4
Exercise:
Problem: / 36
Exercise:
Problem: J 64
Solution:
8
Exercise:
Problem
Exercise:
Problem
:V7121
:V¥144
Solution:
12
Exercise:
Problem
Exercise:
Problem
: 169
2/225
Solution:
is:
Exercise:
Problem:
Exercise:
Problem:
V27
V32
Solution:
2
Exercise:
Problem
Exercise:
Problem
: 0/256
: 7216
Solution:
6
Exercise:
Problem:
Exercise:
Problem
: V400
Solution:
20
Exercise:
Problem
Exercise:
Problem
: V'900
: /10,000
Solution:
100
Exercise:
Problem
: 324
Exercise:
Problem: \/3,600
Solution:
60
For the following problems, use a calculator with the keys ./z, y”, and 1/x
to find each of the values.
Exercise:
Problem: J 676
Exercise:
Problem: \/1,156
Solution:
34
Exercise:
Problem: ,/46,225
Exercise:
Problem: ,/17,288,964
Solution:
4,158
Exercise:
Problem
Exercise:
Problem
: 3/3,375
: /331,776
Solution:
24
Exercise:
Problem
Exercise:
Problem
: °/5,764,801
: 2/16,777,216
Solution:
4
Exercise:
Problem
Exercise:
Problem
: */16,777,216
: 1/9765 ,625
Solution:
3]
Exercise:
Problem: </160,000
Exercise:
Problem: \/531,441
Solution:
81
Exercises for Review
Exercise:
Problem:
({link]) Use the numbers 3, 8, and 9 to illustrate the associative
property of addition.
Exercise:
Problem:
({link]) In the multiplication 8 - 4 = 32, specify the name given to the
numbers 8 and 4.
Solution:
8 is the multiplier; 4 is the multiplicand
Exercise:
Problem: ({link]) Does the quotient 150 exist? If so, what is it?
Exercise:
Problem: ([link]) Does the quotient 0+15exist? If so, what is it?
Solution:
Yes; 0
Exercise:
Problem:
({link]) Use the numbers 4 and 7 to illustrate the commutative property
of multiplication.
Grouping Symbols and the Order of Operations
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This
module discusses grouping symbols and the order of operations. By the end of the module students
should be able to understand the use of grouping symbols, understand and be able to use the order of
operations and use the calculator to determine the value of a numerical expression.
Section Overview
e Grouping Symbols
e Multiple Grouping Symbols
e The Order of Operations
e Calculators
Grouping Symbols
Grouping symbols are used to indicate that a particular collection of numbers and meaningful
operations are to be grouped together and considered as one number. The grouping symbols
commonly used in mathematics are the following:
CLE
Parentheses: ( )
Brackets: [ |
Braces: { }
Bar:
In a computation in which more than one operation is involved, grouping symbols indicate which
operation to perform first. If possible, we perform operations inside grouping symbols first.
Sample Set A
If possible, determine the value of each of the following.
Example:
9+ (3-8)
Since 3 and 8 are within parentheses, they are to be combined first.
9+(8-8) =9+424
= 33
Thus,
9+ (3-8) = 33
Example:
(10+0) -6
Since 10 + 0 is undefined, this operation is meaningless, and we attach no value to it. We write,
"undefined."
Practice Set A
If possible, determine the value of each of the following.
Exercise:
Problem: 16 — (3 - 2)
Solution:
10
Exercise:
Problem: 5 + (7 - 9)
Solution:
68
Exercise:
Problem: (4 + 8) - 2
Solution:
24
Exercise:
Problem: 28~(18 — 11)
Solution:
4
Exercise:
Problem: (33+3) — 11
Solution:
0
Exercise:
Problem: 4 + (0+0)
Solution:
not possible (indeterminant)
Multiple Grouping Symbols
When a set of grouping symbols occurs inside another set of grouping symbols, we perform the
operations within the innermost set first.
Sample Set B
Determine the value of each of the following.
Example:
2+ (8-3) -—(5+6)
Combine 8 and 3 first, then combine 5 and 6.
2+24—11 Nowcombine left to right.
26 — 11
15
Example:
10 + [30 — (2-9)]
Combine 2 and 9 since they occur in the innermost set of parentheses.
10 + [30 — 18] Now combine 30 and 18.
foe te
22,
Practice Set B
Determine the value of each of the following.
Exercise:
Problem: (17 + 8) + (9 + 20)
Solution:
34
Exercise:
Problem: (55 — 6) — (13 - 2)
Solution:
23
Exercise:
Problem: 23 + (12+4) — (11-2)
Solution:
4
Exercise:
Problem: 86 + [14+(10 — 8)|
Solution:
93
Exercise:
Problem: 31 + {9 + [1 + (35 — 2)]}
Solution:
74
Exercise:
Problem: {6 — [24+-(4- 2)]}°
Solution:
27
The Order of Operations
Sometimes there are no grouping symbols indicating which operations to perform first. For example,
suppose we wish to find the value of 3 + 5 - 2. We could do either of two things:
Add 3 and 5, then multiply this sum by 2.
$+5°2 =8-2
= 16
Multiply 5 and 2, then add 3 to this product.
$+5-2 =38+10
= 13
We now have two values for one number. To determine the correct value, we must use the accepted
order of operations.
Order of Operations
1. Perform all operations inside grouping symbols, beginning with the innermost set, in the order 2,
3, 4 described below,
2. Perform all exponential and root operations.
3. Perform all multiplications and divisions, moving left to right.
4. Perform all additions and subtractions, moving left to right.
Sample Set C
Determine the value of each of the following.
Example:
21+ 3-12 Multiply first.
21+ 36 Add.
57
Example:
(15 — 8)+ 5-(6+ 4). Simplify inside parentheses first.
Ta lg Multiply.
7+ 50 Add.
57
Example:
63 — (4+ 6-3)+76—4 Simplify first within the parenthesis by multiplying, then adding.
ie VE Re
G3) PEE FS Is Now perform the additions and subtractions, moving left to right.
aes fi Add 41 and 76: 41 + 76 = 117.
117-4 Subtract 4 from 117: 117 — 4 = 113.
1S.
Example:
7-6—47+1° Evaluate the exponential forms, moving left to right.
7-6—16+1 Multiply 7 and6: 7-6 = 42
a2 Oe Subtract 16 from 42: 42 — 16 = 26
26 + 1 Add 26 and 1: 26 + 1 = 27
27
Example:
6 - (32 + 27) + 4?
6-(9+4)+4
6- (13) + 4?
6- (13) + 16
78 + 16
94
67422 13+8?
4246-22 ' 102-19.5
36+4 a: 1+64
16+6-4 100—19-5
3644 a 1+64
16+24 100—95
40 65
Tay ae Ge
leeds
14
Practice Set C
Evaluate the exponential forms in the parentheses: 3? = 9 and 2? = 4
Add the 9 and 4 in the parentheses: 9+ 4 = 13
Evaluate the exponential form: 4/7 = 16
Multiply 6 and 13: 6-13 = 78
Add 78 and 16: 78 + 16 = 94
Recall that the bar is a grouping symbol.
The fraction 3 is equivalent to(6? + 27)+(4? + 6 - 27)
Determine the value of each of the following.
Exercise:
Problem: 8 + (32 — 7)
Solution:
33
Exercise:
Problem: (34 + 18 — 2-3) +11
Solution:
57
Exercise:
Problem: 8(10)
Solution:
4(2 +3) — (20+ 3-15 + 40-5)
0
Exercise:
Problem: 5 - 8 + 4? — 2?
Solution:
52
Exercise:
Problem: 4(6” — 3°)+(4? — 4)
Solution:
3
Exercise:
Problem: (8 + 9-3)+7+5-(8+4+7+3-5)
Solution:
125
Exercise:
31.93 2194 . 8
Problem: 22" 4 5(£4 )3 8-341
62—29 7232 J * 23-3
Solution:
7
Calculators
Using a calculator is helpful for simplifying computations that involve large numbers.
Sample Set D
Use a calculator to determine each value.
Example:
9,842 + 56-85
Key
Perform the multiplication first. Type
Press
Type
Now perform the addition. Press
Type
Press
The display now reads 14,602.
Example:
42(27 + 18) + 105(810+18)
Key
Operate inside the parentheses Type
Press
Type
Press
Multiply by 42. Press
Type
Press
Place this result into memory by pressing the memory key.
56
9842
27
18
Display Reads
56
56
85
4760
9842
14602
Display Reads
2H
27
18
45
45
42
1890
Now operate in the other parentheses.
Now multiply by 105.
We are now ready to add these two quantities together.
Press the memory recall key.
Thus, 42(27 + 18) + 105(810+18) = 6,615
Example:
16° + 37°
Nonscientific Calculators
Key
Type 16
Press x
Type 16
Press x
Type 16
Press x
Key
Type 810
Press =
Type 18
Press =
Press x
Type 105
Press =
Press a
Press =
Display Reads
16
16
16
256
16
4096
Display Reads
810
810
18
45
45
105
4725
4725
1890
6615
Type
Press
Press the memory key
Type
Press
Type
Press
Type
Press
Press
Press memory recall key
Press
Calculators with y* Key
Key
Type
Press
Type
Press
Press
Type
Press
Type
16
16
Display Reads
16
16
4096
16
65536
37
37
37
1396
37
50653
50653
65536
116189
Press = 116189
Thus, 164 + 37° = 116,189
We can certainly see that the more powerful calculator simplifies computations.
Example:
Nonscientific calculators are unable to handle calculations involving very large numbers.
85612 - 21065
Key Display Reads
Type 85612 85612
Press x 85612
Type 21065 21065
Press =
This number is too big for the display of some calculators and we'll probably get some kind of error
message. On some scientific calculators such large numbers are coped with by placing them in a form
called "scientific notation." Others can do the multiplication directly. (1803416780)
Practice Set D
Use a calculator to find each value.
Exercise:
Problem: 9,285 + 86(49)
Solution:
13,499
Exercise:
Problem: 55(84 — 26) + 120(512 — 488)
Solution:
6,070
Exercise:
Problem: 106° — 17°
Solution:
1,107,495
Exercise:
Problem: 6,053°
Solution:
This number is too big for a nonscientific calculator. A scientific calculator will probably give
you 2.217747109 x10"!
Exercises
For the following problems, find each value. Check each result with a calculator.
Exercise:
Problem: 2 + 3 - (8)
Solution:
26
Exercise:
Problem: 18 + 7 - (4— 1)
Exercise:
Problem: 3 + 8- (6 — 2) +11
Solution:
46
Exercise:
Problem: 1 — 5 - (8 — 8)
Exercise:
Problem: 37 — 1 - 6”
Solution:
1
Exercise:
Problem:
Exercise:
Problem:
Solution:
0
Exercise:
Problem:
Exercise:
Problem:
Solution:
3
Exercise:
Problem:
Exercise:
Problem:
Solution:
26
Exercise:
Problem:
Exercise:
Problem:
Solution:
97
Exercise:
082927
(47 = 224) 9?
/9+14
V100 + V81 — 4?
/848-2-5
V16 145?
61 — 22 + 4[3- (10) + 11]
121 — 4-[(4) - (5) — 12] +
Problem: O48 +5-(12)
Exercise:
Problem:
Solution:
29
Exercise:
8-(6+20) , 3-(64+16)
|
8 l 22
Problem: 10 - [8 + 2- (6 + 7)|
Exercise:
Problem: 21—7~3
Solution:
1
Exercise:
Problem: 10? - 352-3 —2-3
Exercise:
Problem: 85-5 - 5 — 85
Solution:
0
Exercise:
Problem: 2
Exercise:
Problem:
Solution:
90
Exercise:
Problem:
2? .3+42°.(6
26-2. { 0120
Exercise:
Problem: 2 - {(7 + 7) +6: [4- (8+ 2)|}
Solution:
508
Exercise:
Problem: 0 + 10(0) + 15- {4-3+ 1}
Exercise:
Problem: 18 + oe
Solution:
19
Exercise:
Problem: (4 + 7) - (8 — 3)
Exercise:
Problem: (6 + 8) - (5 + 2 — 4)
Solution:
144
Exercise:
Problem: (21 — 3) - (6 — 1) - (7) +. 4(6 + 3)
Exercise:
Problem: (10 + 5) - (10+ 5) — 4- (60 — 4)
Solution:
1
Exercise:
Problem: 6 - {2 - 8 + 3} — (5) -(2)+ $+ (1+8)-(1+11)
Exercise:
Problem: 2° + 3 - (8 + 1)
Solution:
52
Exercise:
Problem:
Exercise:
Problem:
Solution:
25,001
Exercise:
Problem:
Exercise:
Problem:
Solution:
i
25
Exercise:
Problem:
Exercise:
Problem:
Solution:
14
Exercise:
Problem:
Exercise:
Problem:
Solution:
34 424.(1+5)
1° +0 +57. (2+8)*
(7) - (16) — 34 + 2?- (17 +. 3?)
23-7
(1+6)°+2
3-641
67-1
23-3
4342-3
25
a
5 (829-6)
25-7
4
24-5
(2+1)3+23+110 — 15?—[2-5]?
e 55?
0
Exercise:
. 682-10? 18(23+77)
Problem: ——>— + 5 (19) 3
Exercise:
Problem: 2 - {6 + [10° — 6/25) \
Solution:
152
Exercise:
Problem: 181 — 3- (2v36 4 39/64)
Exercise:
2. (vai-¥7125)
Problem: —_P-i10n2?
Solution:
4
5
Exercises for Review
Exercise:
Problem:
({link]) The fact that 0 + any whole number = that particular whole number is an example of
which property of addition?
Exercise:
Problem: ((link]) Find the product. 4,271 x 630.
Solution:
2,690,730
Exercise:
Problem: ({link]) In the statement 27 + 3 = 9, what name is given to the result 9?
Exercise:
Problem: ({link]) What number is the multiplicative identity?
Solution:
1
Exercise:
Problem: ((link]) Find the value of 2+.
Prime Factorization of Natural Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This
module discusses prime factorization of natural numbers. By the end of the module students should
be able to determine the factors of a whole number, distinguish between prime and composite
numbers, be familiar with the fundamental principle of arithmetic and find the prime factorization
of a whole number.
Section Overview
e Factors
e Determining the Factors of a Whole Number
e Prime and Composite Numbers
¢ The Fundamental Principle of Arithmetic
e The Prime Factorization of a Natural Number
Factors
From observations made in the process of multiplication, we have seen that
(factor) - (factor) = product
Factors, Product
The two numbers being multiplied are the factors and the result of the multiplication is the
product. Now, using our knowledge of division, we can see that a first number is a factor of a
second number if the first number divides into the second number a whole number of times
(without a remainder).
One Number as a Factor of Another
A first number is a factor of a second number if the first number divides into the second number a
whole number of times (without a remainder).
We show this in the following examples:
Example:
3 is a factor of 27, since 273 = 9, or 3-9 = 27.
Example:
7 is a factor of 56, since 56+7 = 8, or 7-8 = 56.
Example:
A is not a factor of 10, since 10+4 = 2R2. (There is a remainder.)
Determining the Factors of a Whole Number
We can use the tests for divisibility from [link] to determine all the factors of a whole number.
Sample Set A
Example:
Find all the factors of 24.
Try 1: 24-1 = 24 1 and 24 are factors
Try 2: 24 is even, so 24 is divisible by 2.
24-2 = 12 2 and 12 are factors
Try 3: 2+4=6 and 6 is divisible by 3, so 24 is divisible by 3.
24~-3 = 8 3 and 8 are factors
Try 4: 24+4=6 4 and 6 are factors
Try 5: 24+5 = 4R4 5 is not a factor.
The next number to try is 6, but we already have that 6 is a factor. Once we come upon a factor
that we already have discovered, we can stop.
All the whole number factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.
Practice Set A
Find all the factors of each of the following numbers.
Exercise:
Problem: 6
Solution:
1, 2,3,6
Exercise:
Problem: 12
Solution:
1, 2, 3, 4, 6, 12
Exercise:
Problem: 18
Solution:
1, 2, 3, 6, 9, 18
Exercise:
Problem: 5
Solution:
15
Exercise:
Problem: 10
Solution:
1, 2,5, 10
Exercise:
Problem: 33
Solution:
1, 3, 11, 33
Exercise:
Problem: 19
Solution:
119
Prime and Composite Numbers
Notice that the only factors of 7 are 1 and 7 itself, and that the only factors of 3 are 1 and 3 itself.
However, the number 8 has the factors 1, 2, 4, and 8, and the number 10 has the factors 1, 2, 5, and
10. Thus, we can see that a whole number can have only two factors (itself and 1) and another
whole number can have several factors.
We can use this observation to make a useful classification for whole numbers: prime numbers and
composite numbers.
Prime Number
A whole number (greater than one) whose only factors are itself and 1 is called a prime number.
The Number 1 is Not a Prime Number
The first seven prime numbers are 2, 3, 5, 7, 11, 13, and 17. Notice that the whole number 1 is not
considered to be a prime number, and the whole number 2 is the first prime and the only even
prime number.
Composite Number
A whole number composed of factors other than itself and 1 is called a composite number.
Composite numbers are not prime numbers.
Some composite numbers are 4, 6, 8, 9, 10, 12, and 15.
Sample Set B
Determine which whole numbers are prime and which are composite.
Example:
39. Since 3 divides into 39, the number 39 is composite: 39 + 3 = 13
Example:
47. A few division trials will assure us that 47 is only divisible by 1 and 47. Therefore, 47 is
prime.
Practice Set B
Determine which of the following whole numbers are prime and which are composite.
Exercise:
Problem: 3
Solution:
prime
Exercise:
Problem: 16
Solution:
composite
Exercise:
Problem: 21
Solution:
composite
Exercise:
Problem: 35
Solution:
composite
Exercise:
Problem: 47
Solution:
prime
Exercise:
Problem: 29
Solution:
prime
Exercise:
Problem: 101
Solution:
prime
Exercise:
Problem: 51
Solution:
composite
The Fundamental Principle of Arithmetic
Prime numbers are very useful in the study of mathematics. We will see how they are used in
subsequent sections. We now state the Fundamental Principle of Arithmetic.
Fundamental Principle of Arithmetic
Except for the order of the factors, every natural number other than 1 can be factored in one and
only one way as a product of prime numbers.
Prime Factorization
When a number is factored so that all its factors are prime numbers. the factorization is called the
prime factorization of the number.
The technique of prime factorization is illustrated in the following three examples.
1.10 = 5-2. Both 2 and 5 are primes. Therefore, 2 - 5 is the prime factorization of 10.
2.11. The number 11 is a prime number. Prime factorization applies only to composite numbers.
Thus, 11 has no prime factorization.
3.60 = 2 - 30. The number 30 is not prime: 30 = 2- 15.
60 =2-2-15
The number 15 is not prime: 15 = 3-5
60 =2-2-3-5
We'll use exponents.
60 = 2°+3«5
The numbers 2, 3, and 5 are each prime. Therefore, 2? .3-5 is the prime factorization of 60.
The Prime Factorization of a Natural Number
The following method provides a way of finding the prime factorization of a natural number.
The Method of Finding the Prime Factorization of a Natural Number
1. Divide the number repeatedly by the smallest prime number that will divide into it a whole
number of times (without a remainder).
2. When the prime number used in step 1 no longer divides into the given number without a
remainder, repeat the division process with the next largest prime that divides the given
number.
3. Continue this process until the quotient is smaller than the divisor.
4. The prime factorization of the given number is the product of all these prime divisors. If the
number has no prime divisors, it is a prime number.
We may be able to use some of the tests for divisibility we studied in [link] to help find the primes
that divide the given number.
Sample Set C
Example:
Find the prime factorization of 60.
Since the last digit of 60 is 0, which is even, 60 is divisible by 2. We will repeatedly divide by 2
until we no longer can. We shall divide as follows:
30 is divisible by 2 again.
15 is not divisible by 2, but it is divisible by 3, the next prime.
5 is not divisble by 3, but it is divisible by 5, the next prime.
The quotient 1 is finally smaller than the divisor 5, and the prime factorization of 60 is the product
of these prime divisors.
60 =2-2+-3°5
We use exponents when possible.
CO ages
Example:
Find the prime factorization of 441.
441 is not divisible by 2 since its last digit is not divisible by 2.
441 is divisible by 3 since 4+ 4+ 1 = 9 and 9 is divisible by 3.
3/441
3}147
7149
17
1
147 is divisible by3(1 + 4+ 7 = 12).
49 is not divisible by 3, nor is it divisible by 5. It is divisible by 7.
The quotient 1 is finally smaller than the divisor 7, and the prime factorization of 441 is the
product of these prime divisors.
441=3-3-7-7
Use exponents.
HANS ope
Example:
Find the prime factorization of 31.
31 is not divisible by 2
31 is not divisible by 3
31 is not divisible by 5
31 is not divisible by 7.
Its last digit is not even
ob 2 —15R1
The quotient, 15, is larger than the divisor, 3. Continue.
The digits 3 + 1 = 4, and 4 is not divisible by 3.
oie ee — Oe
The quotient, 10, is larger than the divisor, 3. Continue.
The last digit of 31 is not 0 or 5.
oe — Olu
The quotient, 6, is larger than the divisor, 5. Continue.
Divide by 7.
31+ 7= 4R1
The quotient, 4, is smaller than the divisor, 7.
We can stop the process and conclude that 31 is a prime number.
The number 31 is a prime number
Practice Set C
Find the prime factorization of each whole number.
Exercise:
Problem: 22
Solution:
22:=2- 11
Exercise:
Problem: 40
Solution:
AQ =D? 5
Exercise:
Problem: 48
Solution:
48 = 24.3
Exercise:
Problem: 63
Solution:
63 = 37-7
Exercise:
Problem: 945
Solution:
945 = 3°.5-7
Exercise:
Problem: 1,617
Solution:
1617 = 3-77-11
Exercise:
Problem: 17
Solution:
17 is prime
Exercise:
Problem: 61
Solution:
61 is prime
Exercises
For the following problems, determine the missing factor(s).
Exercise:
14
Problem:
Solution:
2
Exercise:
Problem:
Exercise:
Problem:
Solution:
4
Exercise:
Problem:
Exercise:
Problem:
Solution:
11
Exercise:
Problem:
Exercise:
Problem:
Solution:
3-2
Exercise:
Problem:
Exercise:
20=4-
36 = 9-
42 = 21.
44—4.
38 = 2-
18=3-
28 =2-
Problem: ee .
Solution:
2-3-5
Exercise:
Problem: C= ee .
For the following problems, find all the factors of each of the numbers.
Exercise:
Problem: 16
Solution:
1, 2, 4, 8, 16
Exercise:
Problem: 22
Exercise:
Problem: 56
Solution:
1, 2, 4, 7, 8, 14, 28, 56
Exercise:
Problem: 105
Exercise:
Problem: 220
Solution:
1, 2, 4,5, 10, 11, 20, 22, 44, 55, 110, 220
Exercise:
Problem: 15
Exercise:
Problem: 32
Solution:
1, 2, 4, 8, 16, 32
Exercise:
Problem: 80
Exercise:
Problem: 142
Solution:
1, 2, 71, 142
Exercise:
Problem: 218
For the following problems, determine which of the whole numbers are prime and which are
composite.
Exercise:
Problem: 23
Solution:
prime
Exercise:
Problem: 25
Exercise:
Problem: 27
Solution:
composite
Exercise:
Problem: 2
Exercise:
Problem: 3
Solution:
prime
Exercise:
Problem: 5
Exercise:
Problem: 7
Solution:
prime
Exercise:
Problem: 9
Exercise:
Problem: 11
Solution:
prime
Exercise:
Problem: 34
Exercise:
Problem: 55
Solution:
composite (5 - 11)
Exercise:
Problem: 63
Exercise:
Problem: 1,044
Solution:
composite
Exercise:
Problem: 924
Exercise:
Problem: 339
Solution:
composite
Exercise:
Problem: 103
Exercise:
Problem: 209
Solution:
composite ( 11 - 19)
Exercise:
Problem: 667
Exercise:
Problem: 4,575
Solution:
composite
Exercise:
Problem: 119
For the following problems, find the prime factorization of each of the whole numbers.
Exercise:
Problem: 26
Solution:
2-13
Exercise:
Problem:
Exercise:
Problem:
Solution:
2-33
Exercise:
Problem:
Exercise:
Problem:
Solution:
a
Exercise:
Problem:
Exercise:
Problem:
Solution:
BP. 3x5
Exercise:
Problem:
Exercise:
Problem:
Solution:
38
54
62
56
176
480
819
2,025
ieee
Exercise:
Problem
2 148,225
Exercises For Review
Exercise:
Problem
: ([link]) Round 26,584 to the nearest ten.
Solution:
26,580
Exercise:
Problem
Exercise:
Problem
: ({link]) How much bigger is 106 than 79?
: ([link]) True or false? Zero divided by any nonzero whole number is zero.
Solution:
true
Exercise:
Problem
Exercise:
Problem
: ({link]) Find the quotient. 10,584 + 126.
: ({link]) Find the value of /121 — /81 + 6? + 3.
Solution:
14
The Greatest Common Factor
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses the greatest common factor. By
the end of the module students should be able to find the greatest common
factor of two or more whole numbers.
Section Overview
e The Greatest Common Factor (GCF)
e A Method for Determining the Greatest Common Factor
The Greatest Common Factor (GCF)
Using the method we studied in [link], we could obtain the prime factoriza-
tions of 30 and 42.
Common Factor
We notice that 2 appears as a factor in both numbers, that is, 2 is a common
factor of 30 and 42. We also notice that 3 appears as a factor in both
numbers. Three is also a common factor of 30 and 42.
Greatest Common Factor (GCF)
When considering two or more numbers, it is often useful to know if there
is a largest common factor of the numbers, and if so, what that number is.
The largest common factor of two or more whole numbers is called the
greatest common factor, and is abbreviated by GCF. The greatest
common factor of a collection of whole numbers is useful in working with
fractions (which we will do in [link]).
A Method for Determining the Greatest Common Factor
A straightforward method for determining the GCF of two or more whole
numbers makes use of both the prime factorization of the numbers and
exponents.
Finding the GCF
To find the greatest common factor (GCF) of two or more whole
numbers:
1. Write the prime factorization of each number, using exponents on
repeated factors.
2. Write each base that is common to each of the numbers.
3. To each base listed in step 2, attach the smallest exponent that appears
on it in either of the prime factorizations.
4. The GCF is the product of the numbers found in step 3.
Sample Set A
Find the GCF of the following numbers.
Example:
12 and 18
, 1 eo ia 2 eS
Chg
2. The common bases are 2 and 3.
3. The smallest exponents appearing on 2 and 3 in the prime
factorizations are, respectively, 1 and 1 ( 2'and 37), or 2 and 3.
4. The GCF is the product of these numbers.
23 — 6
The GCF of 30 and 42 is 6 because 6 is the largest number that divides
both 30 and 42 without a remainder.
Example:
18, 60, and 72
1820 = Dee 284
TG ne ye
(GES RG DE I OD Bs UE Fe
2. The common bases are 2 and 3.
3. The smallest exponents appearing on 2 and 3 in the prime
factorizations are, respectively, 1 and 1:
2! from 18
31 from 60
4. The GCF is the product of these numbers.
GCENS? 3 — 6
Thus, 6 is the largest number that divides 18, 60, and 72 without a
remainder.
Example:
700, 1,880, and 6,160
700
eA VS" 2 iis Se ea es
= BoB Gah
= 25 7
eh = BoC) = Bee sAliy = ae BORE
Se Do AT
Ty — 9°.5.47
6,160 = 2-3,080 ee ay =
= 25-711
NI
. The common bases are 2 and 5
i
The smallest exponents appearing on 2 and 5 in the prime
factorizations are, respectively, 2 and 1.
2? from 700.
5 from either 1,880 or 6,160.
4. The GCF is the product of these numbers.
Gris 2? 5.4 5 0)
Thus, 20 is the largest number that divides 700, 1,880, and 6,160 without a
remainder.
Practice Set A
Find the GCF of the following numbers.
Exercise:
Problem: 24 and 36
Solution:
12
Exercise:
Problem
-48-and 72
Solution:
24
Exercise:
Problem
: 50 and 140
Solution:
10
Exercise:
Problem
‘21 and 225
Solution:
3
Exercise:
Problem
: 450, 600, and 540
Solution:
30
Exercises
For the following problems, find the greatest common factor (GCF) of the
numbers.
Exercise:
Problem
:6 and 8
Solution:
2
Exercise:
Problem
Exercise:
Problem
:5 and 10
: 8 and 12
Solution:
4
Exercise:
Problem
Exercise:
Problem
:9 and 12
: 20 and 24
Solution:
4
Exercise:
Problem
Exercise:
235 and 175
Problem:
Solution:
3)
Exercise:
Problem:
Exercise:
Problem:
Solution:
33
Exercise:
Problem:
Exercise:
Problem:
Solution:
9
Exercise:
Problem:
Exercise:
Problem:
25 and 45
45 and 189
66 and 165
264 and 132
99 and 135
65 and 15
33 and 77
Solution:
11
Exercise:
Problem: 245 and 80
Exercise:
Problem: 351 and 165
Solution:
3
Exercise:
Problem: 60, 140, and 100
Exercise:
Problem: 147, 343, and 231
Solution:
7
Exercise:
Problem: 24, 30, and 45
Exercise:
Problem: 175, 225, and 400
Solution:
25
Exercise:
Problem
Exercise:
Problem
: 210, 630, and 182
: 14, 44, and 616
Solution:
2
Exercise:
Problem
Exercise:
Problem
: 1,617, 735, and 429
? 1,573, 4,862, and 3,553
Solution:
11
Exercise:
Problem
Exercise:
Problem
> 3,672, 68, and 920
: 7, 2,401, 343, 16, and 807
Solution:
1
Exercise:
Problem: 500, 77, and 39
Exercise:
Problem: 441, 275, and 221
Solution:
1
Exercises for Review
Exercise:
Problem: ({link]) Find the product. 2,753 x 4,006.
Exercise:
Problem: ({link]) Find the quotient. 954 + 18.
Solution:
53
Exercise:
Problem:
({link]) Specify which of the digits 2, 3, or 4 divide into 9,462.
Exercise:
Problem: ({link]) Write 8x 8x8x8x8x8 using exponents.
Solution:
8© — 262,144
Exercise:
Problem: ([link]) Find the prime factorization of 378.
The Least Common Multiple
This module is from Fundamentals of Mathematics by Denny Burzynski and
Wade Ellis, Jr. This module discusses the least common multiple. By the end
of the module students should be able to find the least common multiple of two
or more whole numbers.
Section Overview
Multiples
Common Multiples
The Least Common Multiple (LCM)
Finding the Least Common Multiple
Multiples
When a whole number is multiplied by other whole numbers, with the
exception of zero, the resulting products are called multiples of the given
whole number. Note that any whole number is a multiple of itself.
Sample Set A
Multiples of Multiples of Multiples of Multiples of
2 3 8 10
2x2 3xl=3 8x1=8 10x1= 10
2X2 A 3x2 =6 8x2 = 16 10x2 = 20
2x3 = 6 3x3 =9 8x3 = 24 10x3 = 30
2x4 = 8 3x4 = 12 8x4 = 32 10x4 = 40
2x5 = 10 3x5 = 15 8x5 = 40
Practice Set A
Find the first five multiples of the following numbers.
Exercise:
Problem: 4
Solution:
4, 8, 12, 16, 20
Exercise:
Problem: 5
Solution:
Oy 10S 1S,.20,25
Exercise:
Problem: 6
Solution:
6, 12, 18, 24, 30
Exercise:
Problem: 7
Solution:
10x5 = 50
TAMA, 215.20. 55
Exercise:
Problem: 9
Solution:
3, 18,27, 36, 45
Common Multiples
There will be times when we are given two or more whole numbers and we
will need to know if there are any multiples that are common to each of them.
If there are, we will need to know what they are. For example, some of the
multiples that are common to 2 and 3 are 6, 12, and 18.
Sample Set B
Example:
We can visualize common multiples using the number line.
Multiples First common Second common Third common
of 2 multiple multiple multiple
Multiples
of 3
Notice that the common multiples can be divided by both whole numbers.
Practice Set B
Find the first five common multiples of the following numbers.
Exercise:
Problem: 2 and 4
Solution:
4, 8, 12, 16, 20
Exercise:
Problem: 3 and 4
Solution:
12, 24, 36, 48, 60
Exercise:
Problem: 2 and 5
Solution:
10, 20, 30, 40, 50
Exercise:
Problem: 3 and 6
Solution:
6, 12; 18, 24, 30
Exercise:
Problem: 4 and 5
Solution:
20, 40, 60, 80, 100
The Least Common Multiple (LCM)
Notice that in our number line visualization of common multiples (above), the
first common multiple is also the smallest, or least common multiple,
abbreviated by LCM.
Least Common Multiple
The least common multiple, LCM, of two or more whole numbers is the
smallest whole number that each of the given numbers will divide into without
a remainder.
The least common multiple will be extremely useful in working with fractions
(link).
Finding the Least Common Multiple
Finding the LCM
To find the LCM of two or more numbers:
1. Write the prime factorization of each number, using exponents on
repeated factors.
2. Write each base that appears in each of the prime factorizations.
3. To each base, attach the largest exponent that appears on it in the prime
factorizations.
4. The LCM is the product of the numbers found in step 3.
There are some major differences between using the processes for obtaining
the GCF and the LCM that we must note carefully:
The Difference Between the Processes for Obtaining the GCF and the
LCM
1. Notice the difference between step 2 for the LCM and step 2 for the GCF.
For the GCF, we use only the bases that are common in the prime
factorizations, whereas for the LCM, we use each base that appears in the
prime factorizations.
2. Notice the difference between step 3 for the LCM and step 3 for the GCF.
For the GCF, we attach the smallest exponents to the common bases,
whereas for the LCM, we attach the largest exponents to the bases.
Sample Set C
Find the LCM of the following numbers.
Example:
9 and 12
eae se
RR 26d
2. The bases that appear in the prime factorizations are 2 and 3.
3. The largest exponents appearing on 2 and 3 in the prime factorizations
are, respectively, 2 and 2:
2? from 12.
3” from 9.
4. The LCM is the product of these numbers.
LCM = 2?-37=4-9 = 36
Thus, 36 is the smallest number that both 9 and 12 divide into without
remainders.
Example:
90 and 630
ie
O(a 2 A oS
O30 0 ied 2 ID oe — io ON
=—9.97%.5.7
2. The bases that appear in the prime factorizations are 2, 3, 5, and 7.
3. The largest exponents that appear on 2, 3, 5, and 7 are, respectively, 1, 2,
lege soralel be
2! from either 90 or 630.
3? from either 90 or 630.
5 from either 90 or 630.
7: from 630.
4. The LCM is the product of these numbers.
eC Mes 3 2 i Oe 630
Thus, 630 is the smallest number that both 90 and 630 divide into with no
remainders.
Example:
33, 110, and 484
So) Seal
CU eee ree ra nN bal
ANE sy es ie I ee
2. The bases that appear in the prime factorizations are 2, 3, 5, and 11.
3. The largest exponents that appear on 2, 3, 5, and 11 are, respectively, 2,
i oles ave’
2? from 484.
3! from 33.
5! from 110
11? from 484.
4. The LCM is the product of these numbers.
LOM = 27.3-5-11?
ees |
7260
Thus, 7260 is the smallest number that 33, 110, and 484 divide into
without remainders.
Practice Set C
Find the LCM of the following numbers.
Exercise:
Problem: 20 and 54
Solution:
540
Exercise:
Problem: 14 and 28
Solution:
28
Exercise:
Problem: 6 and 63
Solution:
126
Exercise:
Problem: 28, 40, and 98
Solution:
1,960
Exercise:
Problem: 16, 27, 125, and 363
Solution:
6,534,000
Exercises
For the following problems, find the least common multiple of the numbers.
Exercise:
Problem: 8 and 12
Solution:
24
Exercise:
Problem: 6 and 15
Exercise:
Problem: 8 and 10
Solution:
40
Exercise:
Problem: 10 and 14
Exercise:
Problem: 4 and 6
Solution:
12
Exercise:
Problem: 6 and 12
Exercise:
Problem: 9 and 18
Solution:
18
Exercise:
Problem: 6 and 8
Exercise:
Problem: 5 and 6
Solution:
30
Exercise:
Problem: 7 and 8
Exercise:
Problem: 3 and 4
Solution:
12
Exercise:
Problem: 2 and 9
Exercise:
Problem:
Solution:
63
Exercise:
Problem:
Exercise:
Problem:
Solution:
72
Exercise:
Problem:
Exercise:
Problem:
Solution:
720
Exercise:
Problem:
Exercise:
Problem:
Solution:
7 and 9
28 and 36
24 and 36
28 and 42
240 and 360
162 and 270
20 and 24
120
Exercise:
Problem: 25 and 30
Exercise:
Problem: 24 and 54
Solution:
216
Exercise:
Problem: 16 and 24
Exercise:
Problem: 36 and 48
Solution:
144
Exercise:
Problem: 24 and 40
Exercise:
Problem: 15 and 21
Solution:
105
Exercise:
Problem: 50 and 140
Exercise:
Problem:
Solution:
231
Exercise:
Problem:
Exercise:
Problem:
Solution:
126
Exercise:
Problem:
Exercise:
Problem:
Solution:
4,410
Exercise:
Problem:
Exercise:
Problem:
7, 11, and 33
8, 10, and 15
18, 21, and 42
4.°5,and 21
45, 63, and 98
15, 25, and 40
12, 16, and 20
Solution:
240
Exercise:
Problem: 84 and 96
Exercise:
Problem: 48 and 54
Solution:
432
Exercise:
Problem: 12, 16, and 24
Exercise:
Problem: 12, 16, 24, and 36
Solution:
144
Exercise:
Problem: 6, 9, 12, and 18
Exercise:
Problem: 8, 14, 28, and 32
Solution:
224
Exercise:
Problem: 18, 80, 108, and 490
Exercise:
Problem: 22, 27, 130, and 225
Solution:
193,050
Exercise:
Problem: 38, 92, 115, and 189
Exercise:
Problem: 8 and 8
Solution:
8
Exercise:
Problem: 12, 12, and 12
Exercise:
Problem: 3, 9, 12, and 3
Solution:
36
Exercises for Review
Exercise:
Problem: (({link]) Round 434,892 to the nearest ten thousand.
Exercise:
Problem: ((link]) How much bigger is 14,061 than 7,509?
Solution:
6,552
Exercise:
Problem: ((link]) Find the quotient. 22,428+14.
Exercise:
Problem: ((link]) Expand 84%. Do not find the value.
Solution:
84 - 84- 84
Exercise:
Problem: ((link]) Find the greatest common factor of 48 and 72.
Summary of Key Concepts
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module reviews the key concepts from the chapter
"Exponents, Roots, Factorization of Whole Numbers."
Summary of Key Concepts
Exponential Notation ((link])
Exponential notation is a description of repeated multiplication.
Exponent ([link])
An exponent records the number of identical factors repeated in a
multiplication.
In a number such as 72,
Base ([link])
7 is called the base.
Exponent ([link])
3 is called the exponent, or power.
Power ((link])
7° is read "seven to the third power," or "seven cubed."
Squared, Cubed ([link])
A number raised to the second power is often called squared. A number
raised to the third power is often called cubed.
Root ({link])
In mathematics, the word root is used to indicate that, through repeated
multiplication, one number is the source of another number.
The Radical Sign ,/" ([link])
The symbol ,/~ is called a radical sign and indicates the square root of a
number. The symbol ,/~ represents the nth root.
Radical, Index, Radicand ({link])
An expression such as ¥/16 is called a radical and 4 is called the index.
The number 16 is called the radicand.
Grouping Symbols ([link])
Grouping symbols are used to indicate that a particular collection of
numbers and meaningful operations are to be grouped together and
considered as one number. The grouping symbols commonly used in
mathematics are
Parentheses: ( )
Brackets: [ |
Braces: { }
Bar:
Order of Operations ({link])
1. Perform all operations inside grouping symbols, beginning with the
innermost set, in the order of 2, 3, and 4 below.
2. Perform all exponential and root operations, moving left to right.
3. Perform all multiplications and division, moving left to right.
4. Perform all additions and subtractions, moving left to right.
One Number as the Factor of Another ([link])
A first number is a factor of a second number if the first number divides
into the second number a whole number of times.
Prime Number ((link])
A whole number greater than one whose only factors are itself and 1 is
called a prime number. The whole number 1 is not a prime number. The
whole number 2 is the first prime number and the only even prime number.
Composite Number ([link])
A whole number greater than one that is composed of factors other than
itself and 1 is called a composite number.
Fundamental Principle of Arithmetic ({link])
Except for the order of factors, every whole number other than 1 can be
written in one and only one way as a product of prime numbers.
Prime Factorization ([link])
The prime factorization of 45 is 3 - 3 - 5. The numbers that occur in this
factorization of 45 are each prime.
Determining the Prime Factorization of a Whole Number ((link])
There is a simple method, based on division by prime numbers, that
produces the prime factorization of a whole number. For example, we
determine the prime factorization of 132 as follows.
The prime factorization of 132 is2-2-3-11 = 2?-3-11.
Common Factor ([(link])
A factor that occurs in each number of a group of numbers is called a
common factor. 3 is a common factor to the group 18, 6, and 45
Greatest Common Factor (GCF) ({link])
The largest common factor of a group of whole numbers is called the
greatest common factor. For example, to find the greatest common factor
of 12 and 20,
1. Write the prime factorization of each number.
12 = 2-2-3=27.3
60 = 2-2-3-5=2987°.3-5
2. Write each base that is common to each of the numbers:
2 and 3
3. The smallest exponent appearing on 2 is 2.
The smallest exponent appearing on 3 is 1.
4. The GCF of 12 and 60 is the product of the numbers 2? and 2.
9?.3=4-3=12
Thus, 12 is the largest number that divides both 12 and 60 without a
remainder.
Finding the GCF ([(link])
There is a simple method, based on prime factorization, that determines the
GCF of a group of whole numbers.
Multiple ({link])
When a whole number is multiplied by all other whole numbers, with the
exception of zero, the resulting individual products are called multiples of
that whole number. Some multiples of 7 are 7, 14, 21, and 28.
Common Multiples ([{link])
Multiples that are common to a group of whole numbers are called
common multiples. Some common multiples of 6 and 9 are 18, 36, and 54.
The LCM ([link])
The least common multiple (LCM) of a group of whole numbers is the
smallest whole number that each of the given whole numbers divides into
without a remainder. The least common multiple of 9 and 6 is 18.
Finding the LCM ({link])
There is a simple method, based on prime factorization, that determines the
LCM of a group of whole numbers. For example, the least common
multiple of 28 and 72 is found in the following way.
1. Write the prime factorization of each number
98 = 2-2.7=2".7
72 = 2-2-9.3-3=29'. 3?
2. Write each base that appears in each of the prime factorizations, 2, 3,
and 7.
3. To each of the bases listed in step 2, attach the Jargest exponent that
appears on it in the prime factorization.
23 32 and 7
4. The LCM is the product of the numbers found in step 3.
93.39%.7=8.9-7=504
Thus, 504 is the smallest number that both 28 and 72 will divide into
without a remainder.
The Difference Between the GCF and the LCM ([link])
The GCF of two or more whole numbers is the largest number that divides
into each of the given whole numbers. The LCM of two or more whole
numbers is the smallest whole number that each of the given numbers
divides into without a remainder.
Exercise Supplement
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is an exercise supplement for the chapter
"Exponents, Roots, Factorization of Whole Numbers" and contains many
exercise problems. Odd problems are accompanied by solutions.
Exercise Supplement
Exponents and Roots ({link])
For problems 1 -25, determine the value of each power and root.
Exercise:
Problem: 3°
Solution:
27
Exercise:
Problem: 4°
Exercise:
Problem: 0°
Solution:
0
Exercise:
Problem: 14
Exercise:
Problem:
Solution:
144
Exercise:
Problem:
Exercise:
Problem:
Solution:
64
Exercise:
Problem:
Exercise:
Problem:
Solution:
a2
Exercise:
Problem:
Exercise:
Problem:
12?
72
82
i
99
34
157
Solution:
225
Exercise:
Problem:
Exercise:
Problem:
25°
Solution:
625
Exercise:
Problem:
Exercise:
Problem
V36
: 225
Solution:
is:
Exercise:
Problem:
Exercise:
Problem:
V64
v16
Solution:
2
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
Exercise:
Problem
Exercise:
Problem
: 7216
» 7144
Solution:
12
Exercise:
Problem
Exercise:
Problem:
: 196
v1
Solution:
1
Exercise:
Problem: </ 0
Exercise:
Problem: </ 64
Solution:
2
Section 3.2
For problems 26-45, use the order of operations to determine each value.
Exercise:
Problem: 2° — 2 - 4
Exercise:
Problem: 5” — 10-2 —5
Solution:
0
Exercise:
Problem: \/81 — 32 + 6-2
Exercise:
Problem: 15° + 52 - 22
Solution:
S20
Exercise:
Problem: 3 - (2? + 37)
Exercise:
Problem: 64 - (2° — 2)
Solution:
64
Exercise:
. Be41 3°)
Problem: ce a 71
Exercise:
. 6-1 4947
Problem: ~~; >
Solution:
as
Exercise:
2-[3+5(2?+1)|
Problem: 5.9332
Exercise:
3?.[2°—14(23+25)|
Problem: a oe
Solution:
9
57
Exercise:
2_.93)\_ 9,
Problem: Aas acs +5- [eps + 1
Exercise:
Problem: (8 — 3)” + (2+ 32)"
Solution:
146
Exercise:
Problem: 32 - (2 sh v2) 423. (vai Z 3°)
Exercise:
Problem: / 16+ 9
Solution:
S)
Exercise:
Problem: / 16 -|- J 9
Exercise:
Problem:
Compare the results of problems 39 and 40. What might we conclude?
Solution:
The sum of square roots is not necessarily equal to the square root of
the sum.
Exercise:
Problem: \/18 - 2
Exercise:
Problem: ‘gy 6-6
Solution:
6
Exercise:
Problem: \/7 - 7
Exercise:
Problem: 4 8-8
Solution:
8
Exercise:
Problem:
An records the number of identical factors that are repeated in a
multiplication.
Prime Factorization of Natural Numbers ({link])
For problems 47- 53, find all the factors of each number.
Exercise:
Problem: 18
Solution:
1.2.3, 6; 9:18
Exercise:
Problem: 24
Exercise:
Problem: 11
Solution:
1,11
Exercise:
Problem: 12
Exercise:
Problem: 51
Solution:
1 Oe an ro
Exercise:
Problem: 25
Exercise:
Problem: 2
Solution:
1,2
Exercise:
Problem: What number is the smallest prime number?
Grouping Symbol and the Order of Operations ({link])
For problems 55 -64, write each number as a product of prime factors.
Exercise:
Problem: 55
Solution:
5-11
Exercise:
Problem: 20
Exercise:
Problem: 80
Solution:
24.5
Exercise:
Problem: 284
Exercise:
Problem: 700
Solution:
22.52.7
Exercise:
Problem: 845
Exercise:
Problem: 1,614
Solution:
2-3-269
Exercise:
Problem: 921
Exercise:
Problem: 29
Solution:
29 is a prime number
Exercise:
Problem: 37
The Greatest Common Factor ((link])
For problems 65 - 75, find the greatest common factor of each collection of
numbers.
Exercise:
Problem
«5 and: 15
Solution:
S
Exercise:
Problem
Exercise:
Problem
: 6 and 14
:10 and 15
Solution:
5
Exercise:
Problem
Exercise:
Problem
+6; 6and 12
: 18 and 24
Solution:
6
Exercise:
Problem
Exercise:
: 42 and 54
Problem: 40 and 60
Solution:
20
Exercise:
Problem: 18, 48, and 72
Exercise:
Problem: 147, 189, and 315
Solution:
21
Exercise:
Problem: 64, 72, and 108
Exercise:
Problem: 275, 297, and 539
Solution:
if
The Least Common Multiple ({link])
For problems 76-86, find the least common multiple of each collection of
numbers.
Exercise:
Problem
Exercise:
Problem
:5and15
: 6 and 14
Solution:
42
Exercise:
Problem
Exercise:
Problem
:10 and 15
: 36 and 90
Solution:
180
Exercise:
Problem
Exercise:
Problem
: 42 and 54
: 8, 12, and 20
Solution:
120
Exercise:
Problem
: 40, 50, and 180
Exercise:
Problem: 135, 147, and 324
Solution:
79, 380
Exercise:
Problem: 108, 144, and 324
Exercise:
Problem: 5, 18, 25, and 30
Solution:
450
Exercise:
Problem: 12, 15, 18, and 20
Exercise:
Problem: Find all divisors of 24.
Solution:
1,253, 4,6, 8,12, 24
Exercise:
Problem: Find all factors of 24.
Exercise:
Problem: Write all divisors of 2°
.52.
Solution:
1, 2, 4, 5, 7, 8, 10, 14, 20, 25, 35, 40, 50, 56, 70, 100, 140, 175, 200,
280, 700, 1,400
Exercise:
Problem: Write all divisors of 6 - 82 - 10°.
Exercise:
Problem: Does 7 divide 5° - 64 - 727 - 8°?
Solution:
yes
Exercise:
Problem: Does 13 divide 8° - 107 - 114 - 13? - 15?
Proficiency Exam
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is a proficiency exam to the chapter
"Exponents, Roots, Factorization of Whole Numbers." Each problem is
accompanied with a reference link pointing back to the module that
discusses the type of problem demonstrated in the question. The problems
in this exam are accompanied by solutions.
Proficiency Exam
Exercise:
Problem:
({link]) In the number 8°, write the names used for the number 8 and
the number 5.
Solution:
base; exponent
Exercise:
Problem:
({link]) Write using exponents. 12 x 12 x 12 x 12 x 12 x 12 x 12
Solution:
12°
Exercise:
Problem: ((link]) Expand 9%.
Solution:
9*=9-9-9-9=6,561
For problems 4-15, determine the value of each expression.
Exercise:
Problem: ({link]) 4°
Solution:
64
Exercise:
Problem: ((link]) 1°
Solution:
1
Exercise:
Problem: ((link]) 0°
Solution:
0
Exercise:
Problem: ({link]) po
Solution:
64
Exercise:
Problem: ({link]) 49
Solution:
7
Exercise:
Problem
: ((link]) 1/27
Solution:
3
Exercise:
Problem
: ([link]) V1
Solution:
1
Exercise:
Problem
: ([link]) 16 + 2 - (8 — 6)
Solution:
20
Exercise:
Problem
: ((link]) 5? — /100 + 8-2—20+5
Solution:
127
Exercise:
Problem
2 9.92 3__4.52
: ([link]) 3- 552* . $545
Solution:
24
Exercise:
Problem: ((link]) ote at
Solution:
8
Exercise:
Problem:
((link}) |(8 — 3)” + (33 — 4/49) | - 2[(10 - 3?) +9] —5
Solution:
5
For problems 16-20, find the prime factorization of each whole number. If
the number is prime, write "prime."
Exercise:
Problem: ({link|) 18
Solution:
3729
Exercise:
Problem: ({link]) 68
Solution:
2? «17
Exercise:
Problem: ({link]) 142
Solution:
2-71
Exercise:
Problem: ({link]) 151
Solution:
prime
Exercise:
Problem: ({link]) 468
Solution:
2737s 13
For problems 21 and 22, find the greatest common factor.
Exercise:
Problem: ([link]) 200 and 36
Solution:
4
Exercise:
Problem: ([link]) 900 and 135
Solution:
45
Exercise:
Problem: ({link]) Write all the factors of 36.
Solution:
1, 2, 3, 4, 6, 9, 12, 18, 36
Exercise:
Problem: ({link]) Write all the divisors of 18.
Solution:
122 3,.0,0518
Exercise:
Problem: ((link]) Does 7 divide into 5? - 6° - 74 - 8? Explain.
Solution:
Yes, because one of the (prime) factors of the number is 7.
Exercise:
Problem: ((link]) Is 3 a factor of 2° - 3? - 5° - 46? Explain.
Solution:
Yes, because it is one of the factors of the number.
Exercise:
Problem: ((link]) Does 13 divide into 11° - 12* - 157? Explain.
Solution:
No, because the prime 13 is not a factor any of the listed factors of the
number.
For problems 28 and 29, find the least common multiple.
Exercise:
Problem: ([link]) 432 and 180
Solution:
2,160
Exercise:
Problem: ([link]) 28, 40, and 95
Solution:
9,320
Objectives
This module contains the learning objectives for the chapter "Introduction
to Fractions and Multiplication and Division of Fractions" from
Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, jr.
After completing this chapter, you should
Fractions of Whole Numbers ({link])
e understand the concept of fractions of whole numbers
e be able to recognize the parts of a fraction
Proper Fractions, improper Fractions, and Mixed Numbers ({link])
¢ be able to distinguish between proper fractions, improper fractions,
and mixed numbers
e be able to convert an improper fraction to a mixed number
e be able to convert a mixed number to an improper fraction
Equivalent Fractions, Reducing Fractions to Lowest Terms, and
Raising Fractions to Higher Terms (({link])
e be able to recognize equivalent fractions
e be able to reduce a fraction to lowest terms
¢ be able to raise a fraction to higher terms
Multiplication of Fractions ({link])
e understand the concept of multiplication of fractions
e be able to multiply one fraction by another
e be able to multiply mixed numbers
e be able to find powers and roots of various fractions
Division of Fractions ({link])
e be able to determine the reciprocal of a number
¢ be able to divide one fraction by another
Applications Involving Fractions ({link]})
¢ be able to solve missing product statements
¢ be able to solve missing factor statements
Fractions of Whole Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses fractions of whole numbers. By
the end of the module students should be able to understand the concept of
fractions of whole numbers and recognize the parts of a fraction.
Section Overview
e More Numbers on the Number Line
e Fractions of Whole Numbers
e The Parts of a Fraction
e Reading and Writing Fractions
More Numbers on the Number Line
In Chapters [link], [link], and [link], we studied the whole numbers and
methods of combining them. We noted that we could visually display the
whole numbers by drawing a number line and placing closed circles at
whole number locations.
0 1 2 3 4 5 6 7 8 9 10 11 12
By observing this number line, we can see that the whole numbers do not
account for every point on the line. What numbers, if any, can be associated
with these points? In this section we will see that many of the points on the
number line, including the points already associated with whole numbers,
can be associated with numbers called fractions.
Fractions of Whole Numbers
The Nature of the Positive Fractions
We can extend our collection of numbers, which now contains only the
whole numbers, by including fractions of whole numbers. We can
determine the nature of these fractions using the number line.
If we place a pencil at some whole number and proceed to travel to the right
to the next whole number, we see that our journey can be broken into
different types of equal parts as shown in the following examples.
a. 1 part.
Whole Next
number whole number
b. 2 equal parts.
Next
whole number
Midway point
c. 3 equal parts.
Pe” AM a, A.
Whole Next
number whole number
d. 4 equal parts.
Whol Next
icribieds whole number
The Latin Word Fractio
Notice that the number of parts, 2, 3, and 4, that we are breaking the
original quantity into is always a nonzero whole number. The idea of
breaking up a whole quantity gives us the word fraction. The word fraction
comes from the Latin word "fractio" which means a breaking, or fracture.
Suppose we break up the interval from some whole number to the next
whole number into five equal parts.
Part Part Part Part Part
1 2 3 4 5
Whole Next
number whole number
After starting to move from one whole number to the next, we decide to
stop after covering only two parts. We have covered 2 parts of 5 equal parts.
This situation is described by writing =
Whol Next
aber whole number
Positive Fraction
2 - ,
A number such as 5 is called a positive fraction, or more simply, a
fraction.
The Parts of a Fraction
A fraction has three parts.
1. The fraction bar — .
Fraction Bar
The fraction bar serves as a grouping symbol. It separates a quantity
into individual groups. These groups have names, as noted in 2 and 3
below.
2. The nonzero number below the fraction bar.
Denominator
This number is called the denominator of the fraction, and it indicates
the number of parts the whole quantity has been divided into. Notice
that the denominator must be a nonzero whole number since the least
number of parts any quantity can have is one.
3. The number above the fraction bar.
Numerator
This number is called the numerator of the fraction, and it indicates
how many of the specified parts are being considered. Notice that the
numerator can be any whole number (including zero) since any
number of the specified parts can be considered.
whole number numerator
ae ee eae a Eg Coa ee Se ee ee
nonzero whole number denominator
Sample Set A
The diagrams in the following problems are illustrations of fractions.
Example:
Diagrams A whole The whole circle divided 1 of the 3
circle into 3 equal parts equal parts
i —f 1 Jor equal parts
3
The fraction ~ is read as "one third."
Example:
A whole The whole rectangle divided into
rectangle 5 equal parts
3 of J equal parts
5
The fraction = "is read as "three fifths."
Example:
I ee
0 1
The number line between 0 and 1
<—____»+_+_}+__+_+-_+_-+—___+
0 1
The number line between 0 and 1 divided
into 7 equal parts
— eee
0 —. . 7 1
6 of the 7 equal parts
3 of the 5
equal parts
ae |
6 —{6| of the equal parts
7
The fraction “ is read as "Six sevenths."
Example:
A whole The whole circle divided into 4 4 of the 4 equal
circle equal parts parts
4 —| 4 lofthe| 4 |equal parts
4
When the numerator and denominator are equal, the fraction represents the
entire quantity, and its value is 1.
nonzero whole number =e)
same nonzero whole number __
Practice Set A
Specify the numerator and denominator of the following fractions.
Exercise:
Problem: -
Solution:
4,7
Exercise:
Cofon
Problem:
Solution:
5.8
Exercise:
. 10
Problem: ie
Solution:
10,15
Exercise:
Problem:
No}
Solution:
19
Exercise:
Problem:
w/o
Solution:
0, 2
Reading and Writing Fractions
In order to properly translate fractions from word form to number form, or
from number form to word form, it is necessary to understand the use of the
hyphen.
Use of the Hyphen
One of the main uses of the hyphen is to tell the reader that two words not
ordinarily joined are to be taken in combination as a unit. Hyphens are
always used for numbers between and including 21 and 99 (except those
ending in zero).
Sample Set B
Write each fraction using whole numbers.
Example:
Fifty three-hundredths. The hyphen joins the words three and hundredths
and tells us to consider them as a single unit. Therefore,
50
fifty three-hundredths translates as 355
Example:
Fifty-three hundredths. The hyphen joins the numbers fifty and three and
tells us to consider them as a single unit. Therefore,
53
fifty-three hundredths translates as +55
Example:
Four hundred seven-thousandths. The hyphen joins the words seven and
thousandths and tells us to consider them as a single unit. Therefore,
four hundred seven-thousandths translates as ae
Example:
Four hundred seven thousandths. The absence of hyphens indicates that the
words seven and thousandths are to be considered individually.
407
four hundred seven thousandths translates as cua)
Write each fraction using words.
Example:
alt 5 ;
3 translates as twenty-one eighty-fifths.
Example:
rain translates as two hundred three-thousandths. A hyphen is needed
between the words three and thousandths to tell the reader that these words
are to be considered as a single unit.
Example:
Re. translates as two hundred three thousandths.
Practice Set B
Write the following fractions using whole numbers.
Exercise:
Problem:one tenth
Solution:
1
10
Exercise:
Problem: eleven fourteenths
Solution:
dd
14
Exercise:
Problem: sixteen thirty-fifths
Solution:
16
35
Exercise:
Problem: eight hundred seven-thousandths
Solution:
800
7,000
Write the following using words.
Exercise:
Problem: ~
Solution:
three eighths
Exercise:
Problem: —
Solution:
one tenth
Exercise:
Problem: =2~
Solution:
three two hundred fiftieths
Exercise:
eae
Problem: 3,190
Solution:
one hundred fourteen three thousand one hundred ninetieths
Name the fraction that describes each shaded portion.
Exercise:
(Ba
Ty
Solution:
3
8
Exercise:
Problem:
Solution:
1
16
In the following 2 problems, state the numerator and denominator, and write
each fraction in words.
Exercise:
Problem:
The number — is used in converting from Fahrenheit to Celsius.
Solution:
5, 9, five ninths
Exercise:
Problem: A dime is _ of a dollar.
Solution:
1, 10, one tenth
Exercises
For the following 10 problems, specify the numerator and denominator in
each fraction.
Exercise:
Problem: a.
Solution:
numerator, 3; denominator, 4
Exercise:
. 9
Problem: a:
Exercise:
Problem: e
Solution:
numerator, 1; denominator, 5
Exercise:
Problem: ~
Exercise:
Problem: -
Solution:
numerator, 7; denominator, 7
Exercise:
Problem: -
Exercise:
. 0
Problem: 7
Solution:
numerator, 0; denominator, 12
Exercise:
4305
Problem: oe
Exercise:
Problem: 18
Solution:
numerator, 18; denominator, 1
Exercise:
. 0
Problem: 6
For the following 10 problems, write the fractions using whole numbers.
Exercise:
Problem: four fifths
Solution:
4
5
Exercise:
Problem: two ninths
Exercise:
Problem
: fifteen twentieths
Solution:
15
20
Exercise:
Problem
Exercise:
: forty-seven eighty-thirds
Problem: ninety-one one hundred sevenths
Solution:
oe
107
Exercise:
Problem
Exercise:
Problem:
Solution:
605
834
Exercise:
Problem
Exercise:
: twenty-two four hundred elevenths
six hundred five eight hundred thirty-fourths
: three thousand three forty-four ten-thousandths
Problem: ninety-two one-millionths
Solution:
92
1,000,000
Exercise:
Problem: one three-billionths
For the following 10 problems, write the fractions using words.
Exercise:
Problem: 2
Solution:
five ninths
Exercise:
6
Problem: a0
Exercise:
Problem: ——
Solution:
eight fifteenths
Exercise:
10
Problem: iz
Exercise:
e 75
Problem: TT
Solution:
seventy-five one hundredths
Exercise:
86
Problem: 735
Exercise:
916
1,014
Problem:
Solution:
nine hundred sixteen one thousand fourteenths
Exercise:
501
10,001
Problem:
Exercise:
18
31,608
Problem:
Solution:
eighteen thirty-one thousand six hundred eighths
Exercise:
. 1
Problem: 00,000
For the following 4 problems, name the fraction corresponding to the
shaded portion.
Exercise:
Problem:
Solution:
A
2
Exercise:
Problem:
/
t—
Exercise:
Problem:
Oa LL
Solution:
4
i
Exercise:
Problem:
| |
For the following 4 problems, shade the portion corresponding to the given
fraction on the given figure.
Exercise:
Problem: ~
[TIT T
Solution:
a | |
Exercise:
Problem: —
Exercise:
Problem: &
Solution:
PEE
Exercise:
Problem: 4
State the numerator and denominator and write in words each of the
fractions appearing in the statements for the following 10 problems.
Exercise:
Problem: A contractor is selling houses on + acre lots.
Solution:
Numerator, 1; denominator, 4; one fourth
Exercise:
Problem:
The fraction a is sometimes used as an approximation to the number
m. (The symbol is read “pi.")
Exercise:
a
Problem: The fraction 3
is used in finding the volume of a sphere.
Solution:
Numerator, 4; denominator, 3; four thirds
Exercise:
Problem: One inch is Wy of a foot.
Exercise:
Problem:
About 2 of the students in a college statistics class received a “B” in
the course.
Solution:
Numerator, 2; denominator, 7; two sevenths
Exercise:
Problem:
The probability of randomly selecting a club when drawing one card
from a standard deck of 52 cards is Be
Exercise:
Problem:
In a box that contains eight computer chips, five are known to be good
and three are known to be defective. If three chips are selected at
random, the probability that all three are defective is ——
Solution:
Numerator, 1; denominator, 56; one fifty-sixth
Exercise:
Problem:
In a room of 25 people, the probability that at least two people have
-. 569
the same birthdate (date and month, not year) is =759-
Exercise:
Problem:
The mean (average) of the numbers 21, 25, 43, and 36 is 18
Solution:
Numerator, 125; denominator, 4; one hundred twenty-five fourths
Exercise:
Problem:
If a rock falls from a height of 20 meters on Jupiter, the rock will be
32 G&
= meters high after > seconds.
Exercises For Review
Exercise:
Problem:
({link]) Use the numbers 3 and 11 to illustrate the commutative
property of addition.
Solution:
3+11=114+3=14
Exercise:
Problem: ({link]) Find the quotient. 676 + 26
Exercise:
Problem: ([link]) Write 7 - 7-7 - 7 - 7 using exponents.
Solution:
79
Exercise:
Problem: ([link]) Find the value of SS + ae
Exercise:
Problem: ({link]) Find the least common multiple of 12, 16, and 18.
Solution:
144
Proper Fractions, Improper Fractions, and Mixed Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses proper fractions, improper
fractions, and mixed numbers. By the end of the module students should be
able to distinguish between proper fractions, improper fractions, and mixed
numbers, convert an improper fraction to a mixed number and convert a
mixed number to an improper fraction.
Section Overview
e Positive Proper Fractions
e Positive Improper Fractions
e Positive Mixed Numbers
e Relating Positive Improper Fractions and Positive Mixed Numbers
e Converting an Improper Fraction to a Mixed Number
e Converting a Mixed Number to an Improper Fraction
Now that we know what positive fractions are, we consider three types of
positive fractions: proper fractions, improper fractions, and mixed numbers.
Positive Proper Fractions
Positive Proper Fraction
Fractions in which the whole number in the numerator is strictly less than
the whole number in the denominator are called positive proper fractions.
On the number line, proper fractions are located in the interval from 0 to 1.
Positive proper fractions are always less than one.
—_—_c?8i— eae
0 1
ee
All proper fractions are located in this interval.
The closed circle at 0 indicates that 0 is included, while the open circle at 1
indicates that 1 is not included.
Some examples of positive proper fractions are
Note that 1 < 2,3 < 5, 20 < 27, and 106 < 225.
Positive Improper Fractions
Positive Improper Fractions
Fractions in which the whole number in the numerator is greater than or
equal to the whole number in the denominator are called positive improper
fractions. On the number line, improper fractions lie to the right of (and
including) 1. Positive improper fractions are always greater than or equal to
dh
a SE ae
0 1
nn
Positive improper fractions
Some examples of positive improper fractions are
Note that 3 > 2,8 > 5,4 > 4, and 105 > 16.
Positive Mixed Numbers
Positive Mixed Numbers
A number of the form
nonzero whole number + proper fraction
is called a positive mixed number. For example, 22 is a mixed number.
On the number line, mixed numbers are located in the interval to the right
of (and including) 1. Mixed numbers are always greater than or equal to 1.
Se
0 1
a re
Positive mixed numbers
Relating Positive Improper Fractions and Positive Mixed
Numbers
A relationship between improper fractions and mixed numbers is suggested
by two facts. The first is that improper fractions and mixed numbers are
located in the same interval on the number line. The second fact, that mixed
numbers are the sum of a natural number and a fraction, can be seen by
making the following observations.
Divide a whole quantity into 3 equal parts.
i
3
col
a
Now, consider the following examples by observing the respective shaded
areas.
In the shaded region, there are 2 one thirds, or 2.
(4) =4
There are 3 one thirds, or = or 1.
3(1) =4 orl
Improper fraction = whole number.
There are 4 one thirds, or +, or 1 and +
4(+) = +or 1 and 5
The terms 1 and i can be represented as 1 + + or 15
Thus,
There are 5 one thirds, or = or 1 and ae
5(4) =2 or 1 and
The terms 1 and $ can be represented as 1 + + or 12.
Thus,
There are 6 one thirds, or 4, or 2.
1 = 6
6(3) =3 =2
Thus,
6 __
3 =2
Improper fraction = whole number.
The following important fact is illustrated in the preceding examples.
Mixed Number = Natural Number + Proper Fraction
Mixed numbers are the sum of a natural number and a proper fraction.
Mixed number = (natural number) + (proper fraction)
For example 15 can be expressed as 1 + — The fraction 5x can be
expressed as 5 + Z
It is important to note that a number such as 5 + _ does not indicate
multiplication. To indicate multiplication, we would need to use a
multiplication symbol (such as -)
Note:5 2 means 5 + - and not 5 - +, which means 5 times - or5
multiplied by =
Thus, mixed numbers may be represented by improper fractions, and
improper fractions may be represented by mixed numbers.
Converting Improper Fractions to Mixed Numbers
To understand how we might convert an improper fraction to a mixed
number, let's consider the fraction, -
|i i
3 | 3 3 | 3
SS
5
1 -
oy = 1 1 i
gS GT gt gt gs
1
= 1 aes
1
= 13
_ ql
Thus, 5 =1-.
We can illustrate a procedure for converting an improper fraction to a mixed
number using this example. However, the conversion is more easily
accomplished by dividing the numerator by the denominator and using the
result to write the mixed number.
Converting an Improper Fraction to a Mixed Number
To convert an improper fraction to a mixed number, divide the numerator
by the denominator.
1. The whole number part of the mixed number is the quotient.
2. The fractional part of the mixed number is the remainder written over
the divisor (the denominator of the improper fraction).
Sample Set A
Convert each improper fraction to its corresponding mixed number.
Example:
2. Divide 5 by 3.
1 <— whole number part
3)5
| 3
2 <— numerator of the fractional part
denominator of the fractional part
The improper fraction = = ie
Example:
* | Divide 46 by 9.
5 <— whole number part
9) 46
45
1 <— numerator of the fractional part
denominator of the fractional part
: : 46 _ fl
The improper fraction | = 5>.
9
5 6
—_——_—————_+>—_+++
0 $58
Example:
$3. Divide 83 by 11.
7 — whole number part
11) 83
17
6 — numerator of the fractional part
denominator of the fractional part
The improper fraction a = 7a.
7 8
tt et
0 Hi, 7
Example:
+ Divide 104 by 4.
26 <— whole number part
4) 104
oO:
24
24
0 — numerator of the fractional part
denominator of the fractional part
iM = 262 = 26
4 4
The improper fraction a 9
25 26 27
0 ter
Practice Set A
Convert each improper fraction to its corresponding mixed number.
Exercise:
Problem: =
Solution:
1
ao:
Exercise:
Problem: oa
Solution:
2
33
Exercise:
14
Problem: I
Solution:
11
Exercise:
Peel
Problem: aa
Solution:
13
Exercise:
Problem: oi
Solution:
3
19 a
Exercise:
Problem: ==
Solution:
62
Converting Mixed Numbers to Improper Fractions
To understand how to convert a mixed number to an improper fraction,
we'll recall
mixed number = (natural number) + (proper fraction)
and consider the following diagram.
2 al Net
3 wa oe 3
——
+ 3
Nee oe eel
1
Recall that multiplication describes repeated addition.
Notice that 3 can be obtained from 12 using multiplication in the
following way.
Multiply: 3-1=3
2
G
Add: 3 + 2 = 5. Place the 5 over the 3: 2
The procedure for converting a mixed number to an improper fraction is
illustrated in this example.
Converting a Mixed Number to an Improper Fraction
To convert a mixed number to an improper fraction,
1. Multiply the denominator of the fractional part of the mixed number
by the whole number part.
2. To this product, add the numerator of the fractional part.
3. Place this result over the denominator of the fractional part.
Sample Set B
Convert each mixed number to an improper fraction.
Example:
BL
8
1. Multiply: 8-5 = 40.
2. Add: 40 + 7 = 47.
3. Place 47 over 8: a
ewes
Thus, 53 =
Example:
2
164
1. Multiply: 3-16 = 48.
2. Add: 48 + 2 = 50.
3, Places UiOver.s, _
Thus, 162 = 2
Practice Set B
Convert each mixed number to its corresponding improper fraction.
Exercise:
Problem: 84
Solution:
33
4
Exercise:
Problem: 5 4
Solution:
28
5
Exercise:
eae
Problem: 155
Solution:
19
15
Exercise:
Problem: 12
“I]t
Solution:
86
7
Exercises
For the following 15 problems, identify each expression as a proper
fraction, an improper fraction, or a mixed number.
Exercise:
Problem: 3
Solution:
improper fraction
Exercise:
Problem:
cols
Exercise:
“or
Problem:
Solution:
proper fraction
Exercise:
Problem: z
Exercise:
Problem: 65
Solution:
mixed number
Exercise:
Problem: —
Exercise:
1,001
12
Problem:
Solution:
improper fraction
Exercise:
Problem: 191
Exercise:
sca
Problem: Lae
Solution:
mixed number
Exercise:
Problem: 312
Exercise:
Problem: 3
Solution:
mixed number
Exercise:
6D:
Problem: iD
Exercise:
Problem: <
Solution:
proper fraction
Exercise:
Problem:
Exercise:
Problem: 101 ;
Solution:
mixed number
For the following 15 problems, convert each of the improper fractions to its
corresponding mixed number.
Exercise:
Problem: —
Exercise:
Problem: =>
Solution:
2
43
Exercise:
Problem: —
Exercise:
Problem:
Solution:
3
87
Exercise:
Problem: —
Exercise:
Problem: ~
Solution:
9
Exercise:
121
Problem: ii
Exercise:
165
Problem: 15
Solution:
13 or 134
Exercise:
Problem: a
Exercise:
Problem:
Solution:
b)
5552
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
364
Exercise:
Problem:
Exercise:
Problem:
Solution:
29
(ore
Exercise:
5,000
316
Problem: 7
For the following 15 problems, convert each of the mixed numbers to its
corresponding improper fraction.
Exercise:
Problem: 4.
Solution:
35
8
Exercise:
2125
Problem: 155
Exercise:
Problem: 64
Solution:
61
9
Exercise:
Problem: 15>
Exercise:
Problem: 10 ~
Solution:
115
11
Exercise:
; 3
Problem: 15 Til
Exercise:
Problem:
Solution:
26
3
Exercise:
Problem:
Exercise:
Problem:
Solution:
107
5
Exercise:
Problem:
Exercise:
Problem:
Solution:
209
21
Exercise:
Problem:
Exercise:
(oe)
eo|bo
aN
ENS (oC)
Problem: 90 Ww
Solution:
9001
100
Exercise:
Problem: 30073,
Exercise:
Problem: 192
Solution:
159
8
Exercise:
Problem: Why does 0 + not qualify as a mixed number?
Note:See the definition of a mixed number.
Exercise:
Problem: Why does 5 qualify as a mixed number?
Note:See the definition of a mixed number.
Solution:
... because it may be written as 5 a , where 7 is any positive whole
number.
Calculator Problems
For the following 8 problems, use a calculator to convert each mixed
number to its corresponding improper fraction.
Exercise:
Problem: 35 +4
Exercise:
Problem: 27 —
Solution:
1,652
61
Exercise:
. 929 40
Problem: 8377
Exercise:
; 21
Problem: 105 ol
Solution:
2,436
23
Exercise:
- 605
Problem: 72 606
Exercise:
Problem: 8162
Solution:
20,419
25
Exercise:
Problem: 708 ==
Exercise:
4,216
Problem: 6,012 57
Solution:
48,803,620
8,117
Exercises For Review
Exercise:
Problem: ([link]) Round 2,614,000 to the nearest thousand.
Exercise:
Problem: ({link]) Find the product. 1,004 - 1,005.
Solution:
1,009,020
Exercise:
Problem: ({link]) Determine if 41,826 is divisible by 2 and 3.
Exercise:
Problem: ({link]) Find the least common multiple of 28 and 36.
Solution:
252
Exercise:
Problem:
12
([link]) Specify the numerator and denominator of the fraction +>.
Equivalent Fractions, Reducing Fractions to Lowest Terms, and Raising
Fractions to Higher Terms
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses equivalent fractions, reducing
fractions to lowest terms, and raising fractions to higher terms. By the end
of the module students should be able to recognize equivalent fractions,
reduce a fraction to lowest terms and be able to raise a fraction to higher
terms.
Section Overview
e Equivalent Fractions
e Reducing Fractions to Lowest Terms
e Raising Fractions to Higher Terms
Equivalent Fractions
Let's examine the following two diagrams.
l
I
=e 4 of the whole is shaded.
Notice that both - and + represent the same part of the whole, that is, they
represent the same number.
Equivalent Fractions
Fractions that have the same value are called equivalent fractions. Equiva-
lent fractions may look different, but they are still the same point on the
number line.
There is an interesting property that equivalent fractions satisfy.
2 4
376
A Test for Equivalent Fractions Using the Cross Product
These pairs of products are called cross products.
2°623°4
12412
If the cross products are equal, the fractions are equivalent. If the cross
products are not equal, the fractions are not equivalent.
Thus, and Ha are equivalent, that is, 4 = -.
Sample Set A
Determine if the following pairs of fractions are equivalent.
Example:
Sand£. Test for equality of the cross products.
The cross products are equals.
The fractions = and — are equivalent, so = = 4.
Example:
oe Test for equality of the cross products.
The cross products are not equal.
The fractions 4 and Je are not equivalent.
Practice Set A
Determine if the pairs of fractions are equivalent.
Exercise:
ok os
Problem: mse
Solution:
6 46,
, yes
Exercise:
, yes
Exercise:
ey, 2S
Problem: =, =
Solution:
30 # 24, no
Exercise:
Problem: =
Solution:
40 ~ 40.
, yes
Exercise:
noc ee
Problem: Re
Solution:
12 ~ 12,
, yes
Reducing Fractions to Lowest Terms
It is often very useful to convert one fraction to an equivalent fraction that
has reduced values in the numerator and denominator. We can suggest a
method for doing so by considering the equivalent eee — and 2.
Pea divide both the numerator and denominator of 2 5 by 3. "The cbs
= and 3 = are equivalent.
(Can you prove this?) So =. We hers to convert ~ to 4 . Now
, 5 a
divide the numerator and ae Ghanaian of 2 = by 3, and see tt happens.
The fraction + is converted to 2,
5
A natural question is "Why did we choose to divide by 3?" Notice that
9 3-3
We can see that the factor 3 is common to both the numerator and
denominator.
Reducing a Fraction
From these observations we can suggest the following method for
converting one fraction to an equivalent fraction that has reduced values in
the numerator and denominator. The method is called reducing a fraction.
A fraction can be reduced by dividing both the numerator and denominator
by the same nonzero whole number.
9 _ 9+38_3 16_16+2_ 8
12 12+3 4 30 30-2, 15
Notice that 3 =1 and >= 2
Consider the collection of equivalent fractions
2D. fa Ya
20° 16° 12°
Reduced to Lowest Terms
Notice that each of the first four fractions can be reduced to the last
fraction, + by dividing both the numerator and denominator by,
respectively, 5, 4, 3, and 2. When a fraction is converted to the fraction that
has the smallest numerator and denominator in its collection of equivalent
fractions, it is said to be reduced to lowest terms. The fractions + 3, 2
and — are all reduced to lowest terms.
Observe a very important property of a fraction that has been reduced to
lowest terms. The only whole number that divides both the numerator and
denominator without a remainder is the number 1. When 1 is the only
whole number that divides two whole numbers, the two whole numbers are
said to be relatively prime.
Relatively Prime
A fraction is reduced to lowest terms if its numerator and denominator are
relatively prime.
Methods of Reducing Fractions to Lowest Terms
Method 1: Dividing Out Common Primes
1. Write the numerator and denominator as a product of primes.
2. Divide the numerator and denominator by each of the common prime
factors. We often indicate this division by drawing a slanted line
through each divided out factor. This process is also called cancelling
common factors.
3. The product of the remaining factors in the numerator and the product
of remaining factors of the denominator are relatively prime, and this
fraction is reduced to lowest terms.
Sample Set B
Reduce each fraction to lowest terms.
a 1 ° °
i se 1 and 3 are relatively prime.
1
ae FEES = = 4 and 5 are relatively prime.
Example:
ieee
6. = Feet ae = je 7 and 13 are relatively prime (and also truly
ial sal
prime)
Example:
il 1
315 B35R 15
336 22:22. B 7 ~ 16 15 and 16 are relatively prime.
Example:
- = aa8 No common prime factors, so 8 and 15 are relatively prime.
The fraction 3. is reduced to lowest terms.
Practice Set B
Reduce each fraction to lowest terms.
Exercise:
Oo]
Problem:
Solution:
a
2
Exercise:
Problem:
=
BI
Solution:
2
5
Exercise:
Problem: —>
Solution:
uk
8
Exercise:
21
Problem: ae
Solution:
ag
16
Exercise:
mye
Problem: 2B
Solution:
19
7
Exercise:
135
Problem: 33
Solution:
b)
9
Method 2: Dividing Out Common Factors
1. Mentally divide the numerator and the denominator by a factor that is
common to each. Write the quotient above the original number.
2. Continue this process until the numerator and denominator are
relatively prime.
Sample Set C
Reduce each fraction to lowest terms.
Example:
2.5 divides into both 25 and 30.
5
& = 7 5 and 6 are relatively prime.
6
Example:
3. Both numbers are even so we can divide by 2.
Ee
< Now, both 9 and 12 are divisible by 3.
= 4 3 and 4 are relatively prime.
RRR ER
Example:
Sa ee Peet
36 = 28 54 ;- 3 and 8 are relatively prime.
Practice Set C
Reduce each fraction to lowest terms.
Exercise:
Az
Problem: i6
Solution:
3
4
Exercise:
goes
Problem: 5A
Solution:
3
8
Exercise:
goad.
Problem: Ti
Solution:
al
4
Exercise:
. 48
Problem: a
Solution:
3
4
Exercise:
. 63
Problem: a
Solution:
co|NI
Exercise:
150
Problem: 540
Solution:
a:
8
Raising Fractions to Higher Terms
Equally as important as reducing fractions is raising fractions to higher
terms. Raising a fraction to higher terms is the process of constructing an
equivalent fraction that has higher values in the numerator and denominator
than the original fraction.
The fractions - and _— are equivalent, that is, - = — Notice also,
3-3 9
5.3 ‘15
Notice that 3 = 1 and that 3 ca a We are not changing the value of 3
From these observations we can suggest the following method for
converting one fraction to an equivalent fraction that has higher values in
the numerator and denominator. This method is called raising a fraction to
higher terms.
Raising a Fraction to Higher Terms
A fraction can be raised to an equivalent fraction that has higher terms in
the numerator and denominator by multiplying both the numerator and
denominator by the same nonzero whole number.
The fraction 4 can be raised to “ by multiplying both the numerator and
denominator by 8.
| that 2 =1.
Most often, we will want to convert a given fraction to an equivalent
fraction with a higher specified denominator. For example, we may wish to
convert 2 to an equivalent fraction that has denominator 32, that is,
This is possible to do because we know the process. We must multiply both
the numerator and denominator of 2 by the same nonzero whole number in
order to 8 obtain an equivalent fraction.
We have some information. The denominator 8 was raised to 32 by
multiplying it by some nonzero whole number. Division will give us the
proper factor. Divide the original denominator into the new denominator.
32+8=4
Now, multiply the numerator 5 by 4.
5+ 4 = 20
Thus,
5 54 20
8 BA < 35
Sample Set D
Determine the missing numerator or denominator.
Example:
- = =. Divide the original denominator into the new denominator.
35+7 = 5 The quotient is 5. Multiply the original numerator by 5.
ee eee als ne ;
+ = Fe = 35 The missing numerator is 15.
Example:
~ = 2. Divide the original numerator into the new numerator.
45+5 = 9 The quotient is 9. Multiply the original denominator by 9.
~ = — = — The missing denominator is 45.
Practice Set D
Determine the missing numerator or denominator.
Exercise:
Pc: ee a
Problem: = = 40
Solution:
32
Exercise:
Problem:
aloo
Solution:
12
Exercise:
Problem: - i
Solution:
4
Exercise:
Problem: — = ->
Solution:
150
Exercise:
Problem: iz = ie
Solution:
88
Exercises
For the following problems, determine if the pairs of fractions are
equivalent.
Exercise:
Problem:
Solution:
equivalent
Exercise:
Problem: 38:
Exercise:
. 5 10
Problem: =; , 5;
Solution:
equivalent
Exercise:
Problem: _
Exercise:
Problem: -
Solution:
not equivalent
Exercise:
.i 7
Problem: = , 75
Exercise:
16 49
25 ? 75
Problem:
Solution:
not equivalent
Exercise:
dD. _20
Problem: 987712
Exercise:
3 _36_
Problem: 55,735
Solution:
not equivalent
Exercise:
oi. lS.
Problem: <7 , 35
Exercise:
3. 1s
Problem: = ,=;
Solution:
equivalent
Exercise:
. 10 15
Problem: =; 5;
Exercise:
oo
Problem: ~~
Solution:
not equivalent
Exercise:
Solution:
not equivalent
For the following problems, determine the missing numerator or
denominator.
Exercise:
cole
—_
bo
Problem:
Exercise:
Problem:
otf
Solution:
6
Exercise:
eo|bo
CO]+
Problem:
Exercise:
Problem:
Solution:
12
Exercise:
Problem:
Exercise:
Problem:
Solution:
12
Exercise:
Problem:
Exercise:
Problem:
Solution:
20
Exercise:
Problem:
Exercise:
one
TN)
| 09
Dor
ons
bole
co|-~
00
Problem:
Solution:
79
Exercise:
Problem:
Exercise:
Problem:
Solution:
48
Exercise:
Problem:
Exercise:
Problem:
Solution:
80
Exercise:
Problem:
Exercise:
bo|eo
colon
Go|
on]
we
Problem: —
Solution:
18
Exercise:
Problem:
Exercise:
Problem:
Solution:
154
Exercise:
Problem: =
Exercise:
Problem:
Solution:
1,472
Exercise:
Problem:
Exercise:
16 96
15 _ 225
160.—~—~C—“‘
a1 2220
12 ~——«168
9 _ ?
13 ~—-.286
Otte ue
33 —s«1518
19 _ 1045
20 ?
. 37 _ 1369
Problem: = =
Solution:
1,850
For the following problems, reduce, if possible, each of the fractions to
lowest terms.
Exercise:
Problem: 2
Exercise:
rae
Problem: 10
Solution:
A
5
Exercise:
ee ea
Problem: i0
Exercise:
Pe
Problem: iA
Solution:
3
7
Exercise:
Problem: —-
Exercise:
Problem:
Solution:
2
7
Exercise:
Problem:
Exercise:
Problem:
Solution:
2
3
Exercise:
Problem:
Exercise:
Problem:
Solution:
B)
D
Exercise:
Problem:
Exercise:
o>
>|
18
14
|
Problem:
Solution:
yy
3
Exercise:
Problem: —
Exercise:
Problem: —
Solution:
va
3
Exercise:
Problem:
Exercise:
Problem:
Solution:
e
35
Exercise:
Problem:
Exercise:
10
12
16
70
40
60
Problem:
Solution:
5
3
Exercise:
Problem:
Exercise:
Problem:
Solution:
18
5
Exercise:
Problem:
Exercise:
Problem:
Solution:
2
3
Exercise:
Problem:
Exercise:
20
12
32
36
10
36
60
12
18
18
27
Problem:
Solution:
3
4
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
2
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
3
Exercise:
Problem:
Exercise:
18
24
32
11
22
27
17
51
16
42
. 39
Problem: a3
Solution:
3
Exercise:
Problem: —
Exercise:
Problem: 22
Solution:
2
Exercise:
Problem:
Exercise:
, 15
Problem: 1G
Solution:
already reduced
Exercise:
15
Problem: os
Exercise:
Problem:
Solution:
9
25
Exercise:
Problem:
Exercise:
Problem:
Solution:
2
b)
Exercise:
Problem:
Exercise:
Problem:
Solution:
9
8
Exercise:
Problem:
Exercise:
36
100
45
32
30
75
72
64
30
105
Problem:
Solution:
23.
30
Exercise:
Problem:
Exercise:
Problem:
Solution:
20
9
Exercise:
Problem:
Exercise:
Problem:
Solution:
i
3
Exercise:
Problem: —
Exercise:
46.
60
75.
45
40
18
108
Problem: 2>
Solution:
az
4
Exercise:
Problem: =-~
Exercise:
. 51
Problem: 7
Solution:
17
18
Exercise:
Problem:
A ream of paper contains 500 sheets. What fraction of a ream of paper
is 200 sheets? Be sure to reduce.
Exercise:
Problem:
There are 24 hours in a day. What fraction of a day is 14 hours?
Solution:
7
12
Exercise:
Problem:
A full box contains 80 calculators. How many calculators are in t ofa
box?
Exercise:
Problem:
There are 48 plants per flat. How many plants are there in 3 of a flat?
Solution:
16
Exercise:
Problem:
A person making $18,000 per year must pay $3,960 in income tax.
What fraction of this person's yearly salary goes to the IRS?
For the following problems, find the mistake.
Exercise:
58s a et
Problem: m4 8 = 0
Solution:
Should be ~ ; the cancellation is division, so the numerator should be
1,
Exercise:
g _ 2+6
oR os 22 /6¢ =. 3
Problem: = ie a
10 +8
Exercise:
las a
Problem: a Yu 8
Solution:
Cancel factors only, not addends; = is already reduced.
Exercise:
.6 _ At _
Problem: o>
Exercise:
Problem: s 9_9Q
Solution:
Same as [link]; answer is + or 1.
Exercises for Review
Exercise:
Problem: ({link]) Round 816 to the nearest thousand.
Exercise:
Problem: ({link]) Perform the division: 0 + 6.
Solution:
0
Exercise:
Problem
Exercise:
Problem
: (Llink]) Find all the factors of 24.
: (Llink]) Find the greatest common factor of 12 and 18.
Solution:
6
Exercise:
Problem
: (Llink]) Convert 2 to a mixed number.
Multiplication of Fractions
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses multiplication of fractions. By the
end of the module students should be able to understand the concept of
multiplication of fractions, multiply one fraction by another, multiply mixed
numbers and find powers and roots of various fractions.
Section Overview
e Fractions of Fractions
e Multiplication of Fractions
e Multiplication of Fractions by Dividing Out Common Factors
e Multiplication of Mixed Numbers
e Powers and Roots of Fractions
Fractions of Fractions
We know that a fraction represents a part of a whole quantity. For example,
two fifths of one unit can be represented by
HEBER
2 of the whole is shaded.
A natural question is, what is a fractional part of a fractional quantity, or,
what is a fraction of a fraction? For example, what $ of 5 ?
We can suggest an answer to this question by using a picture to examine =
1
of x
é ’ 1
First, let’s represent =.
1
2
nine
5 of the whole is shaded.
Then divide each of the 5 parts into 3 equal parts.
Each part is - of the whole.
Now we’ll take + of the > unit.
2p eer 2: j ea
A of me ¢> Which reduces to me
Multiplication of Fractions
Now we ask, what arithmetic operation (+, —, x, +) will produce “ from 2
: 3
Notice that, if in the fractions and +, we multiply the numerators
together and the denominators together, we get precisely =
2-1
21 _ 2
32 6
This reduces to - as before.
Using this observation, we can suggest the following:
1. The Word "OF" Indicates Multiplication The word "of" translates
to the arithmetic operation "times."
2. The Method of Multiplying Fractions To multiply two or more
fractions, multiply the numerators together and then multiply the
denominators together. Reduce if necessary.
numerator] | _numerator2 J _ _numeratorl . _ numerator 2
denominator 1 denominator 2 denominator 1 denominator 2
Sample Set A
Perform the following multiplications.
Example:
Oe Ol te Oe ee
oe ae ee NOW, PeGuce:
1
eye eerie
ee
Thus
a ol 1
A hG = a8
This means that A of - is = that is, of < of a unit is z of the original
unit.
Example:
5 ; é Bk 4
= +4. Write 4 as a fraction by writing +
3
ig eet os See 0 tee CO ee
goon 8-1 8 y 2
2
oe eee
Woe a
This means that - of 4 whole units is 3 of one whole unit.
Example:
Digi oe
Baas
16
This means that = of = of = of a whole unit is a of the original unit.
Practice Set A
Perform the following multiplications.
Exercise:
Problem:
orp
ony
Solution:
=s
15
Exercise:
Problem:
|e
<o|00
Solution:
2
9
Exercise:
e 4 . 15
Problem: reat
Solution:
2
12
Exercise:
Problem: (+) (+)
Solution:
CO]
Exercise:
Problem: (L ) (2)
Solution:
14
by)
Exercise:
Dor
oo|~I
Problem:
Solution:
35.
48
Exercise:
TN)
Ol
Problem:
Solution:
10
3
Exercise:
Problem: (3) (10)
Solution:
15
2
Exercise:
Problem: = - 8. .
4 aes
Solution:
b)
18
Multiplying Fractions by Dividing Out Common Factors
We have seen that to multiply two fractions together, we multiply
numerators together, then denominators together, then reduce to lowest
terms, if necessary. The reduction can be tedious if the numbers in the
fractions are large. For example,
oe ee oe
9 10 9-10 90
16° 21 ~ 1621 ~ 336 168 28
We avoid the process of reducing if we divide out common factors before
we multiply.
Divide 3 into 9 and 21, and divide 2 into 10 and 16. The product is a
fraction that is reduced to lowest terms.
The Process of Multiplication by Dividing Out Common Factors
To multiply fractions by dividing out common factors, divide out factors
that are common to both a numerator and a denominator. The factor being
divided out can appear in any numerator and any denominator.
Sample Set B
Perform the following multiplications.
Example:
ee
5 66
2 1
IT ee Nine 2:
B g ~~ ties
1 3
Divide 4 and 6 by 2
Divide 5 and 5 by 5
Example
Bae ol
TER sai
4 Z
1 St eee. Ves
im i = 3s eS
3 5
Divide 8 and 10 by 2.
Divide 8 and 12 by 4.
Example:
a|N
ice!
9 39 12
1
1 ff
Uae ee eee ee
9 sf yw 916 ~ 54
x 6
1
Practice Set B
Perform the following multiplications.
Exercise:
TN)
co|~I
Problem:
Solution:
ats
12
Exercise:
e 25 . 10
Problem: io ae
Solution:
25
54
Exercise:
40 | 72
Problem: i on
Solution:
2
3
Exercise:
Problem: 7 - *«
Solution:
TS)
Exercise:
Problem: 12 -
Coco
Solution:
9
2
Exercise:
Problem: (2 ) ( — )
Solution:
1
Exercise:
Gs g220" ,, BI.
Problem: 0. ee ad
Solution:
14
b)
Multiplication of Mixed Numbers
Multiplying Mixed Numbers
To perform a multiplication in which there are mixed numbers, it is
convenient to first convert each mixed number to an improper fraction, then
multiply.
Sample Set C
Perform the following multiplications. Convert improper fractions to mixed
numbers.
Example
Le AD
ie z 4-
Convert each mixed number to an improper fraction.
(a Seal =
“Se Saeee.:
gees ES aaa eel
So een
3 7
Ee ey ed ee
ye: 1 4 4
4 1
Example:
16-85
Convert 8= to an improper fraction.
ole ras co Beil ee lk
i ae AR
aia
ihaeg? aie
There are no common factors to divide out.
th, al LO ob pa
5 5
1 5 1-5
Example:
1 3
1 oa 1
Convert to improper fractions.
gf =) bent a os
GG
Practice Set C
Perform the following multiplications. Convert improper fractions to mixed
numbers.
Exercise:
sow 1
Problem: ae . 27
Solution:
6
Exercise:
ee 3
Problem: 64 . 3a5
Solution:
22
Exercise:
Problem: 7 - 12
Solution:
i
85 =
Exercise:
92 3
Problem: 2= : 35 -3
oc)
Solution:
30
Powers and Roots of Fractions
Sample Set D
Find the value of each of the following.
Example:
(Ayal. touid
Ve See Ge ae
Example:
/ — . We’re looking for a number, call it ?, such that when it is squared,
325 is produced.
2 9
(?)" = apg
We know that
3? = 9 and 10” = 100
We’ |l try +. Since
( 3 ve ashe 8 3-3 9
10 10 10 ‘10-10 100
fee
10am . 10
Example
2 100
45° 4/ qi
wu
a TE ae
a ae
i th
2 sO
45 pi 4
Practice Set D
Find the value of each of the following.
Exercise:
2
Problem: ( ~ )
Solution:
as
64
Exercise:
Problem: (3 ) :
Solution:
9
100
Exercise:
Problem:
Solution:
2
3
Exercise:
Problem:
Solution:
i
2
Exercise:
Problem:
Solution:
i
8
Exercise:
Problem:
Solution:
2
85
Exercise:
co]
Al
Pa) ey ee led
Problem: 2 3 16
Solution:
8
wie
Exercises
For the following six problems, use the diagrams to find each of the
following parts. Use multiplication to verify your result.
Exercise:
- 3 ni
Problem: me of 7
Solution:
i
4
Exercise:
a. 3
Problem: a of =
Exercise:
28 4
Problem: 7 of -
Solution:
i
4
Exercise:
Exercise:
det 1
Problem: z of =
Exercise:
ee 6
Problem: a of =
For the following problems, find each part without using a diagram.
Exercise:
fal 4
Problem: = of =
Solution:
2
5
Exercise:
Problem:
Exercise:
Problem:
Solution:
2
9
Exercise:
Problem:
Exercise:
Problem:
Solution:
A
15
Exercise:
Problem:
Exercise:
Problem:
Solution:
3
5
onloo
®|R
<|00
Exercise:
Problem:
Exercise:
Problem:
Solution:
2
Exercise:
Problem: 2
Exercise:
Problem:
Solution:
10
1
3 or 33
Exercise:
Problem:
Exercise:
Problem:
Solution:
a2
Exercise:
18 38
i9 Of 54
of 2
DOr
IN)
Problem: 142 of 84
Exercise:
e 8 3 2
Problem: , of of a
Solution:
CO]
Exercise:
eed? 6 26:
Problem: : of ia of 3G
Exercise:
Fae nee a eee ot
Problem: 5 of : of ri
Solution:
—
24
Exercise:
Problem: 1dof 5 of 85
Exercise:
cig 5 5
Problem: 2~ of 5 a of 7 z
Solution:
126
For the following problems, find the products. Be sure to reduce.
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
4
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
3
Exercise:
Problem:
Exercise:
Problem:
Solution:
£
9
Exercise:
co]
bole
| 00
TN)
bole
o| co
lor
<|00
14
"15
Problem:
Exercise:
Problem:
Solution:
1
Exercise:
Problem: —-
Exercise:
Problem:
Solution:
28
33
Exercise:
Problem:
Exercise:
Problem:
Solution:
4
3
Exercise:
“Ne
a
|e
35
36
21
76
99
z|N
—_
oo
20
48
55
15
38
Problem: mo ag,
Exercise:
Problem: — -
Solution:
12
Exercise:
Problem:
Exercise:
Problem:
Solution:
13 160
737 Olay
Exercise:
Problem:
Exercise:
Problem:
Solution:
12
Exercise:
M4 |
15
eo |O0
18
14
onloo
21
- 20
_ 16
21
Problem: - .
Exercise:
4. £05
Problem: A
Solution:
18
Exercise:
» 18
Problem: ig
Exercise:
Problem: ~
Solution:
25 1
or 85
Exercise:
Problem: >
Exercise:
Problem: 5 -
Solution:
cl eee pe
8 =! 8
Exercise:
- 33
- 38
- 10
0 |eo
Problem: 16 -
Exercise:
Problem:
eo|bo
Solution:
6
Exercise:
Problem:
G0|eo
Exercise:
Problem: —-
Solution:
ae ee
g=15
Exercise:
Problem: == -
Exercise:
Problem: b=
Solution:
tat)
Exercise:
oS Lae
0
32
[oo
TN)
or
Problem: 22 53
Exercise:
Problem: 64 QA
Solution:
Exercise:
Problem: 95 - > - 1
Exercise:
eee 13 1
Problem: 35 etl
Solution:
de
Exercise:
Problem: 207 - 82 - 162
Exercise:
Problem: (4 ) :
Solution:
an) FING
Exercise:
Problem: (2 ) :
Exercise:
Problem: (+ ) :
Solution:
eae
121
Exercise:
Problem: (= ) :
Exercise:
Problem: (s ) :
Solution:
i
4
Exercise:
Problem: (2. ee.
Exercise:
Problem: (4)? oe
Solution:
=o
15
Exercise:
Problem: (+ ) :
co|00
Exercise:
Problem: (1)” - (2)°
onto
Solution:
=o
25
Exercise:
Problem: (2) - (+)
For the following problems, find each value. Reduce answers to lowest
terms or convert to mixed numbers.
Exercise:
Problem: / 5
Solution:
2
3
Exercise:
; 16
Problem: a5
Exercise:
Problem: , / 2
121
Solution:
9
iG
Exercise:
: 36
Problem: 19
Exercise:
, | 144
Problem: >
Solution:
Exercise:
Problem: ee Ara
Exercise:
Problem: = -
Solution:
1
3
Exercise:
Problem: eg ---
Exercise:
Problem: (1 ae 7
Solution:
7
8
Exercise:
2
Problem: (2) 36 ia
Exercises for Review
Exercise:
Problem: ({link]) How many thousands in 342,810?
Solution:
2
Exercise:
Problem: ({link]) Find the sum of 22, 42, and 101.
Exercise:
Problem: ([link]) Is 634,281 divisible by 3?
Solution:
yes
Exercise:
Problem: ({link]) Is the whole number 51 prime or composite?
Exercise:
Problem: ({link]) Reduce a to lowest terms.
Solution:
6_
25
Division of Fractions
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses division of fractions. By the end
of the module students should be able to determine the reciprocal of a
number and divide one fraction by another.
Section Overview
e Reciprocals
e Dividing Fractions
Reciprocals
Reciprocals
Two numbers whose product is 1 are called reciprocals of each other.
Sample Set A
The following pairs of numbers are reciprocals.
Example:
3 4A
vand3
Notice that we can find the reciprocal of a nonzero number in fractional
form by inverting it (exchanging positions of the numerator and
denominator).
Practice Set A
Find the reciprocal of each number.
Exercise:
Problem: io
Solution:
10
3
Exercise:
eo|po
Problem:
Solution:
3
2
Exercise:
Problem:
oo|—I
Solution:
~1|00
Exercise:
Problem: $
Solution:
5
Exercise:
Problem: 2 4
Note: Write this number as an improper fraction first.
Solution:
a
16
Exercise:
Problem: 5+
Solution:
4
21
Exercise:
Problem: 10 —
Solution:
16
163
Dividing Fractions
Our concept of division is that it indicates how many times one quantity is
contained in another quantity. For example, using the diagram we can see
that there are 6 one-thirds in 2.
There are 6 one-thirds in 2.
Since 2 contains six 3's we express this as
a+|i|me
3
Note also that 2- | 3 |=6
_
: and 8 are reciprocals
Using these observations, we can suggest the following method for dividing
a number by a fraction.
Dividing One Fraction by Another Fraction
To divide a first fraction by a second, nonzero fraction, multiply the first
traction by the reciprocal of the second fraction.
Invert and Multiply
This method is commonly referred to as "invert the divisor and
multiply."
Sample Set B
Perform the following divisions.
Example:
ees RAN OS rea 1 4
3 ~ 7- The divisor is +. Its reciprocal is 3. Multiply = by =
PLLA Le Pere eS
Be ee)
Bleeding See
SEES)
Example:
32 8 ES ages -. 4 eres 4
ras 4 The divisor is +. Its reciprocal is =. Multiply = by =.
ag Sere Slee
2 10
pereneen
ee Cane
Pe as
2 se pe) Weve ul aber See Ih eral mover Ss 2 Multiply 2 by 4
il ie
zB PL ee Oe y)
Ye ae OU
1 1
ee ee
Example
Digest) Oe 0
92 — = 2.
9 9
gi 3-341 10
3 ae ee eae
a ee Dhediniconts ” . Its reciprocal is =>. Multiply > aie ae
Example;
+2 ae First conveniently write 8 as >
+ ea ® The divisor is 8 . Its ear 1S = +. Multiply #3 oer by =
3
pa eee a pees
(ge a OP
2
DS meni eng
i =
Example
Tamm eg 83
= op a5: Ue divisor is... Its reciprocal is >
1
1 Bal
VG ae Saree ete pale
a ae 2-1-7 14
2 5s 7
1
DA oI sae yal
S20
Example:
How many 2 3 -inch-wide packages can be placed in a box 19 inches wide?
The problem is to determine how many two and three eighths are contained
in 19, that is, what is 19 + 22?
22 = a Convert the divisor 2 3. to an improper fraction.
Ihe i Write the egos 19 as a
= + 19 The divisor is 2. Its are are
it
PD p canes nly Pod Bin 8
1 MW “1 il
1
Thus, 8 packages will fit into the box.
Practice Set B
Perform the following divisions.
Exercise:
a ae:
Problem: a
Solution:
CO]
Exercise:
Problem: = — =
Solution:
1
Exercise:
ea er te 5
Problem: is > IB
Solution:
i
2
Exercise:
oi 8
Problem: 8 — ip
Solution:
15
Exercise:
Problem: 65 + -
Solution:
15
Exercise:
Problem: 34 a 12
Solution:
2
Exercise:
Qe De 8s
Problem: Borg oe
Solution:
ea
5
Exercise:
Problem:
A container will hold 106 ounces of grape juice. How many 62-ounce
glasses of grape juice can be served from this container?
Solution:
16 glasses
Determine each of the following quotients and then write a rule for this type
of division.
Exercise:
Problem: 1 — ~
Solution:
bo|eo
Exercise:
Problem: 1 — A
Solution:
e2|00
Exercise:
Problem: 1 — 2
Solution:
a
3
Exercise:
Problem: 1 — -
Solution:
2
5
Exercise:
Problem: When dividing 1 by a fraction, the quotient is the .
Solution:
is the reciprocal of the fraction.
Exercises
For the following problems, find the reciprocal of each number.
Exercise:
ons
Problem:
Solution:
5 1
4 or 15
Exercise:
Problem:
—
[20
Exercise:
Problem: a
Solution:
9
i
5 or 45
Exercise:
Problem:
onl
Exercise:
Ale
Problem: 3
Solution:
ames
13
Exercise:
Problem: 84
Exercise:
Problem: 34
Solution:
7
23
Exercise:
Problem: 5 +
Exercise:
Problem: 1
Solution:
1
Exercise:
Problem: 4
For the following problems, find each value.
Exercise:
. 3. 3
Problem: Roe
Solution:
5
8
Exercise:
Problem: 2
Exercise:
Problem: —
Solution:
=e
10
Exercise:
Problem: $ +
Exercise:
Problem: = —
Solution:
225 29
i96 OF Lig
Exercise:
Problem:
Exercise:
Problem:
Solution:
3
by)
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
Exercise:
Problem:
Exercise:
Problem: —~
Solution:
49
100
Exercise:
2=+0
hey Be ake
ae
3
0+ 5
4. 4
Pele te * iE.
2.2
5 rs
Ti, 2 10
10° 7%
Problem:
Exercise:
Problem:
Solution:
3
5
Exercise:
Problem:
Exercise:
Problem:
Solution:
6
7
Exercise:
Problem:
Exercise:
Problem:
Solution:
4 6
223
1 1
1. 1
1. 1
2 6
35 > op
85 oh
7 or 14%
Exercise:
eek
° 6
fon) To
Problem: 5
Exercise:
Problem: * + 34
Solution:
28 _ 14 a
i Sg or i 9
Exercise:
Problem: 55 a is
Exercise:
Problem: 83 = c
Solution:
10
Exercise:
ae
Problem: — + 15
Exercise:
og ES.
Problem: 3 + =
Solution:
10
1
: or 3=
Exercise:
Problem: 11 is
Exercise:
Problem:
Solution:
A
Fi
Exercise:
Problem:
Exercise:
Problem:
Solution:
3 1
oy or Le
Exercise:
Problem:
Exercise:
Problem:
Solution:
4
5
Exercise:
: b)
=. oS
Be ug, Aa
22-912
16 oe
a. oe 6 5
3 56
455 + las
Ais ait
1000 100
3. 9 6
8 ° 16 5
i. 2. BS
Problem: eee
Exercise:
Pe. a ee
Problem: ce
Solution:
3
Exercise:
4h 4d 3
Problem: a0 1 a a0
Exercise:
Problem: 83 SO
Solution:
1
Exercises for Review
Exercise:
Problem: ({link]) What is the value of 5 in the number 504,216?
Exercise:
Problem: ({link]) Find the product of 2,010 and 160.
Solution:
321,600
Exercise:
Problem:
({link]) Use the numbers 8 and 5 to illustrate the commutative property
of multiplication.
Exercise:
Problem: ({link]) Find the least common multiple of 6, 16, and 72.
Solution:
144
Exercise:
Problem: ((Jink]) Find $ of 6.
Applications Involving Fractions
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses applications involving fractions.
By the end of the module students should be able to solve missing product
statements and solve missing factor statements.
Section Overview
e Multiplication Statements
e Missing Product Statements
e Missing Factor Statements
Multiplication Statements
Statement, Multiplication Statement
A statement is a sentence that is either true or false. A mathematical
statement of the form
product = (factor 1) (factor 2)
is a multiplication statement. Depending on the numbers that are used, it
can be either true or false.
Omitting exactly one of the three numbers in the statement will produce
exactly one of the following three problems. For convenience, we'll
represent the omitted (or missing) number with the letter M (M for
Missing).
1. M = (factor 1) -: (factor 2) Missing product statement.
2.M - (factor 2) = product Missing factor statement.
3. (factor 1) : M = product Missing factor statement.
We are interested in developing and working with methods to determine the
missing number that makes the statement true. Fundamental to these
methods is the ability to translate two words to mathematical symbols. The
word
of translates to times
is translates to equals
Missing Products Statements
The equation M = 8 - 4 is a missing product statement. We can find the
value of M that makes this statement true by multiplying the known factors.
Missing product statements can be used to determine the answer to a
question such as, "What number is fraction 1 of fraction 2?
Sample Set A
Find a of 5 We are being asked the question, "What number is oh of al
We must translate from words to mathematical symbols.
What number is : of 2 becomes
Cg einen
a oe ee
3 8 :
M aS 9 Multiply.
missing known = known
product factor factor
Wi it
a a.
3 i eae
Thus, 7 of Meee
What number is : of 24
Seemmesiain ommmrnoomenaast
tst4
3
= <= 24
a 4
missing known known
product factor factor
6
M = +.. Af — 36 — 8 = 18
Bi,
‘i
Thus, 18 is + of 24.
Practice Set A
Exercise:
Problem: Find 3 of i.
Solution:
2
5
Exercise:
Problem: What number is _ of ~ ?
Solution:
3
4
Exercise:
Problem: —- of _ is what number?
Solution:
oa
6
Missing Factor Statements
The equation 8 - M = 32 is a missing factor statement. We can find the
value of M that makes this statement true by dividing (since we know that
32 +8 = 4).
8-M=832 means that M = 32 + 8
I db bod 8
missing = product + known
factor factor
Finding the Missing Factor
To find the missing factor in a missing factor statement, divide the product
by the known factor.
missing factor = (product) + (known factor)
Missing factor statements can be used to answer such questions as
1. 4 of what number is 3?
2. What part of 12 is 143?
Sample Set B
“ of what number is =?
———
1 od el
3 9
8 ~ 4
known missing product
factor factor
Now, using
missing factor = (product) + (known factor)
We get
3 2
i ae eee ee ee a2
sama ares ae ek ae a
1 1
32
Tl
— 6
Check: Lee ed
4
3
3.¢,9
S14
4
3:3,9
rr Tea
9,9
a
3 -. 9
Thus,3 Of 6.18 7.
2 13
What part of 1- is1—?
ee pot! 14
7
oe ae |
2 13
M * 1— = i—
7 14
missing known product
factor factor
For convenience, let's convert the mixed numbers to improper fractions.
Cea
M: 7 aA
Now, using
missing factor = (product)+(known factor)
we get
14 14° 9
25) vail
2:1
on
= 19
3 927
Check: 2 714
3-9, 27
2-7" 14
27. 27
14 "14
Practice Set B
Exercise:
Problem: 3 of what number is oa?
Solution:
3
4
Exercise:
Problem: 34 of what number is 2 +?
Solution:
16
27
Exercise:
Problem: What part of 2 is --?
Solution:
1
tlh
Exercise:
10°
Problem: What part of 1+ isl t?
Solution:
1
tlh
Exercises
Exercise:
Problem: Find
Solution:
1
>
Exercise:
Problem: Find
Exercise:
Problem: Find
Solution:
Ad
3
Exercise:
Problem:
Exercise:
Problem:
Solution:
9 ra
3 or 1=
Exercise:
Problem:
Exercise:
Problem:
Solution:
=
16
Exercise:
Problem:
Exercise:
Problem:
Solution:
3
4
a of 4B is what number?
14 of 20 ; 2
is of 57 1S what number’
— of _ is what number?
_ of 2 is what number?
+ of 3 is what number?
Exercise:
fee
Problem: TT of is what number?
100
Exercise:
Problem: 0 of 5 is what number?
Solution:
aS
1,000
Exercise:
Problem: 1 2 of 24 is what number?
Exercise:
Problem: 15 of i is what number?
Solution:
10
27
Exercise:
Problem: 15 of 1 is what number?
Exercise:
en 1 9
Problem: Find = of a of 5
Solution:
1
2
Exercise:
Problem:
Exercise:
Problem:
Solution:
2
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
6
Exercise:
Problem:
Exercise:
Problem:
Solution:
2
3
Exercise:
- 4 5 9 4
Find Obs, of a
a of what number is 2?
ios of what number is 2?
» of what number is 2?
— of what number is 2?
~ of what number is 1 ~ ?
Problem:
Exercise:
Problem:
Solution:
1
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
Exercise:
Problem:
Exercise:
Problem:
Solution:
3 2
; or 13
Exercise:
z of what number is x?
- of what number is +?
6
t of what number is 3?
- of what number is &?
bi
- of what number is 0?
= of what number is 1?
Problem
Exercise:
Problem:
: 3= of what number is 1?
a “5 of what number is 5 +?
Solution:
3
Exercise:
Problem
Exercise:
Problem
oak, <9 89
3 35 of what number is 2 is!
: What part of ~ is 15?
Solution:
5 2
3 orl
Exercise:
Problem
Exercise:
Problem
‘ 9 5693
: What part of 55 is 32?
: What part of is =o
Solution:
o7
40
Exercise:
: DA oie ke,
Problem: What part of += is 35?
Exercise:
Problem: What part of 3 is x?
Solution:
=.
15
Exercise:
Problem: What part of 8 is 2?
Exercise:
Problem: What part of 24 is 9?
Solution:
3
8
Exercise:
Problem: What part of 42 is 26?
Exercise:
Problem: Find — of 2
Solution:
ve
10
Exercise:
Problem: = of x is what number?
Exercise:
Problem: « of what number is 22?
Solution:
ns pee, Be
5 a 2
Exercise:
AT 225
Problem: 1 of what number is Ta
Exercise:
Problem: - of what number is 1?
Solution:
16 5
che or Lor
Exercise:
Problem: What part of - is 334?
Exercise:
Problem: = of 3S is what number?
Solution:
30
77
Exercises for Review
Exercise:
Problem:
({link]) Use the numbers 2 and 7 to illustrate the commutative property
of addition.
Exercise:
Problem: (({link]) Is 4 divisible by 0?
Solution:
no
Exercise:
Problem: ({link]) Expand 3”. Do not find the actual value.
Exercise:
Problem: ({link]) Convert 32. to an improper fraction.
Solution:
aL
12
Exercise:
ite ee! ee
Problem: ((link]) Find the value of = + 75: =.
Summary of Key Concepts
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module reviews the key concepts from the chapter
"Introduction to Fractions and Multiplication and Division of Fractions."
Summary of Key Concepts
Fraction ([{link])
The idea of breaking up a whole quantity into equal parts gives us the word
fraction.
Fraction Bar, Denominator, Numerator ((link])
A fraction has three parts:
1. The fraction bar — -
2. The nonzero whole number below the fraction bar is the denominator.
3. The whole number above the fraction bar is the numerator.
oe numerator
— <— fraction bar
“~~ denominator
Proper Fraction ([link])
Proper fractions are fractions in which the numerator is strictly less than
the denominator.
~ is a proper fraction
Improper Fraction ([link])
Improper fractions are fractions in which the numerator is greater than or
equal to the denominator. Also, any nonzero number placed over 1 is an
improper fraction.
a a and = are improper fractions
4°? 5 il
Mixed Number ([{link])
A mixed number is a number that is the sum of a whole number and a
proper fraction.
3 : —_ it
1 is a mixed number (1z —1+ =)
Correspondence Between Improper Fractions and Mixed Numbers
({link])
Each improper fraction corresponds to a particular mixed number, and each
mixed number corresponds to a particular improper fraction.
Converting an Improper Fraction to a Mixed Number ([link])
A method, based on division, converts an improper fraction to an equivalent
mixed number.
4 can be converted to 1 +
Converting a Mixed Number to an Improper Fraction ([link])
A method, based on multiplication, converts a mixed number to an
equivalent improper fraction.
5 i can be converted to a
Equivalent Fractions ({link])
Fractions that represent the same quantity are equivalent fractions.
+ and ° are equivalent fractions
Test for Equivalent Fractions ((link])
If the cross products of two fractions are equal, then the two fractions are
equivalent.
sa 28
47S8
3-824-6
24 = 24
Thus, 4 and ~ are equivalent.
Relatively Prime ({link])
Two whole numbers are relatively prime when 1 is the only number that
divides both of them.
3 and 4 are relatively prime
Reduced to Lowest Terms ([link])
A fraction is reduced to lowest terms if its numerator and denominator are
relatively prime.
The number 3 is reduced to lowest terms, since 3 and 4 are relatively
prime.
The number 2 is not reduced to lowest terms since 6 and 8 are not
relatively prime.
Reducing Fractions to Lowest Terms ({link])
Two methods, one based on dividing out common primes and one based on
dividing out any common factors, are available for reducing a fraction to
lowest terms.
Raising Fractions to Higher Terms ((link])
A fraction can be raised to higher terms by multiplying both the numerator
and denominator by the same nonzero number.
Be 2 eG.
4 4-2 8
The Word “OF” Means Multiplication ({link])
In many mathematical applications, the word "of" means multiplication.
Multiplication of Fractions ({link])
To multiply two or more fractions, multiply the numerators together and
multiply the denominators together. Reduce if possible.
4 5-4 20 1
“15 8-15 120 £46
olen
Multiplying Fractions by Dividing Out Common Factors ([link])
Two or more fractions can be multiplied by first dividing out common
factors and then using the rule for multiplying fractions.
1
23 - &
1 1
1. 1
awa
B®
2 3
Multiplication of Mixed Numbers ((link])
To perform a multiplication in which there are mixed numbers, first convert
each mixed number to an improper fraction, then multiply. This idea also
applies to division of mixed numbers.
Reciprocals ({link])
Two numbers whose product is 1 are reciprocals.
7 and > are reciprocals
Division of Fractions ({link])
To divide one fraction by another fraction, multiply the dividend by the
reciprocal of the divisor.
ons
2 2
1D
Dividing 1 by a Fraction ([link])
When dividing 1 by a fraction, the quotient is the reciprocal of the fraction.
Multiplication Statements ({link])
A mathematical statement of the form
product = (factor 1) (factor 2)
is a multiplication statement.
By omitting one of the three numbers, one of three following problems
result:
1. M = (factor 1) « (factor 2) Missing product statement.
2. product = (factor 1) : M Missing factor statement.
3. product = M : (factor 2) Missing factor statement.
Missing products are determined by simply multiplying the known factors.
Missing factors are determined by
missing factor = (product) + (known factor)
Exercise Supplement
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is an exercise supplement for the chapter
"Introduction to Fractions and Multiplication and Division of Fractions"
and contains many exercise problems. Odd problems are accompanied by
solutions.
Exercise Supplement
Fractions of Whole Numbers ([link])
For Problems 1 and 2, name the suggested fraction.
Exercise:
Problem:
Solution:
oT)
1
or 3
Exercise:
Problem:
A
For problems 3-5, specify the numerator and denominator.
Exercise:
Problem: <
Solution:
numerator, 4; denominator, 5
Exercise:
ns)
Problem: ao
Exercise:
Problem: z
Solution:
numerator, 1; denominator, 3
For problems 6-10, write each fraction using digits.
Exercise:
Problem: Three fifths
Exercise:
Problem: Eight elevenths
Solution:
8
ti
Exercise:
Problem: Sixty-one forty firsts
Exercise:
Problem: Two hundred six-thousandths
Solution:
200
6,000
Exercise:
Problem: zero tenths
For problems 11-15, write each fraction using words.
Exercise:
10
Problem: i7
Solution:
ten seventeenths
Exercise:
21
Problem: 35
Exercise:
606
Problem: Ta
Solution:
six hundred six, one thousand four hundred thirty-firsts
Exercise:
Problem: ~
Exercise:
aii
Problem: ae
Solution:
one sixteenth
For problems 16-18, state each numerator and denominator and write each
fraction using digits.
Exercise:
Problem: One minute is one sixtieth of an hour.
Exercise:
Problem:
In a box that contains forty-five electronic components, eight are
known to be defective. If three components are chosen at random from
the box, the probability that all three are defective is fifty-six fourteen
thousand one hundred ninetieths.
Solution:
numerator, 56; denominator, 14,190
Exercise:
Problem:
About three fifths of the students in a college algebra class received a
“B” in the course.
For problems 19 and 20, shade the region corresponding to the given
fraction.
Exercise:
Problem: -
|i tT
Solution:
“iil
Exercise:
Problem: ~
LTT EL
Proper Fraction, Improper Fraction, and Mixed Numbers ((link])
For problems 21-29, convert each improper fraction to a mixed number.
Exercise:
Problem: r.
Solution:
3
27
Exercise:
Problem:
Exercise:
Problem:
Solution:
3
6%
Exercise:
Problem:
Exercise:
On
oo|
121
Problem: ——
Solution:
2
1182
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
lz
Exercise:
Problem:
Exercise:
to |co
eon
Problem: +
Solution:
3
For problems 30-40, convert each mixed number to an improper fraction.
Exercise:
Problem: 5 4
Exercise:
Problem: 16+
Solution:
129
8
Exercise:
Problem: 18=
Exercise:
Problem: 3=
Solution:
16
5
Exercise:
ond
Problem: 255
Exercise:
Problem: 17 a
Solution:
377
21
Exercise:
Problem: 1 <
Exercise:
Problem: 1 5
Solution:
3
2
Exercise:
Problem: 25
Exercise:
Problem: 82
Solution:
62
7
Exercise:
Problem: 22
Exercise:
Problem: Why does 0 Ta not qualify as a mixed number?
Solution:
because the whole number part is zero
Exercise:
Problem: Why does 8 qualify as a mixed number?
Equivalent Fractions, Reducing Fractions to Lowest Terms, and
Raising Fractions to Higher Term ([link])
For problems 43-47, determine if the pairs of fractions are equivalent.
Exercise:
ed 2
Problem: =, 35
Solution:
equivalent
Exercise:
~ 8 32
Problem: >, 33
Exercise:
e 3 24
Problem: 57, 7
Solution:
not equivalent
Exercise:
93 38
Problem: 25 16
Exercise:
oe AOS. “qd:
Problem: 7» 1 3
Solution:
not equivalent
For problems 48-60, reduce, if possible, each fraction.
Exercise:
. 10
Problem: oe
Exercise:
a8
Problem: re
Solution:
2
11
Exercise:
Problem: 2°22
Exercise:
Problem: =
Solution:
5
11
Exercise:
Problem:
Exercise:
Problem:
Solution:
3
5
Exercise:
Problem: —
Exercise:
Problem:
Solution:
2
iN
Exercise:
Problem:
Exercise:
Problem:
Solution:
35
68
Exercise:
18
25
21
35
45
85
24
136
182
Problem: = 30
Exercise:
325
Problem: S10
Solution:
65
162
Exercise:
250
Problem: 000
For problems 61-72, determine the missing numerator or denominator.
Exercise:
Problem: 3 =—- +>
35
Solution:
15
Exercise:
ion. ee
Problem: ii = 30
Exercise:
es eee
Problem: 2 7
Solution:
6
Exercise:
Problem:
Exercise:
Problem:
Solution:
27
Exercise:
Problem: ——
Exercise:
Problem: =
Solution:
4?
Exercise:
Problem:
Exercise:
Problem:
Solution:
168
Exercise:
SS — 25
se
il _ 33
Gt #
ees
15 —?
ia,
15 ~~ 45
aaa
a> 20
AD c= 190"
21 °”~—s?
Problem: == =
Exercise:
Problem: =
15 180
a
Solution:
192
Exercise:
Problem: a=
21 336
?
Multiplication and Division of Fractions ({link], [link])
For problems 73-95, perform each multiplication and division.
Exercise:
i424,
Problem: St Ge
Solution:
3
4
Exercise:
e 8 e 3
Problem: a. ae
Exercise:
Problem: — -
Solution:
ae
24
Exercise:
Problem:
Exercise:
Problem:
Solution:
b)
36
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
Exercise:
Problem:
Exercise:
Problem:
Solution:
wT
15 )
Ba, aot
6 22 39
2.15 5
3 7 6
1 7
3749
dt
257
a ae
15 16 24
1
48
Exercise:
Problem: —
Exercise:
Problem:
Solution:
ae
35
Exercise:
Problem:
Exercise:
Problem:
Solution:
50 _ wl
= a
Exercise:
Problem:
Exercise:
Problem:
Solution:
2
8 . 3
1p + 35
14 . 8
1p 7 39
3
18-53
3 1
3222.
1 4
1 1
6 230
10
; OT
Exercise:
62.55 2 9 7
Problem: 3 Tae 2 i:
Exercise:
Problem: 7 — 24
Solution:
3
Exercise:
Problem: 17 — 44
Exercise:
Problem: 2. -- 14
Solution:
Alt
2
Exercise:
Problem: 2
eo] bo
oN)
| oo
Exercise:
Problem: 20 - a
Solution:
90
Exercise:
Problem: 0 — 4+
Exercise:
2 ae gel. 25
Problem: 1 = 65 a's
Solution:
1
Applications Involving Fractions ({link])
Exercise:
Problem: Find - of =,
Exercise:
219
Problem: What part of 3 aT
Solution:
‘4 1
= OF om
Exercise:
Problem: What part of 34 is 155?
Exercise:
ae 2 9
Problem: Find 65 of 7°
Solution:
4
Exercise:
14.5
Problem: a» of what number is Te
Exercise:
Problem: What part of 4% is 33?
Solution:
12
13
Exercise:
Problem: Find 8-4 of 162.
Exercise:
° 3 .. 185
Problem: i of what number is 39!
Solution:
4
Exercise:
Problem: Find ~ of 0.
Exercise:
Problem: Find — of 1.
Solution:
A
12
Proficiency Exam
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is a proficiency exam to the chapter
"Introduction to Fractions and Multiplication and Division of Fractions. '
Each problem is accompanied with a reference link pointing back to the
module that discusses the type of problem demonstrated in the question.
The problems in this exam are accompanied by solutions.
Proficiency Exam
Exercise:
Problem: ({link]) Shade a portion that corresponds to the fraction 2.
TTT
Solution:
Exercise:
Problem:
({link]) Specify the numerator and denominator of the fraction 2.
Solution:
Numerator, 5; denominator, 9
Exercise:
Problem: ({link]) Write the fraction five elevenths.
Solution:
se
11
Exercise:
Problem: ({link]) Write, in words, -.
Solution:
Four fifths
Exercise:
Problem:
({link]) Which of the fractions is a proper fraction? 4, =, =
Solution:
b)
12
Exercise:
Problem: ({link]) Convert 34 to an improper fraction.
Solution:
25
7
Exercise:
Problem: ({link]) Convert ~- to a mixed number.
Solution:
3
oe
Exercise:
Problem: ({link]) Determine if +s and 3 are equivalent fractions.
Solution:
yes
For problems 9-11, reduce, if possible, each fraction to lowest terms.
Exercise:
Problem: ({link]) oo
Solution:
3
5
Exercise:
Problem: ((Jink]) =
Solution:
a.
7
Exercise:
Problem: (({link]) —
Solution:
13
60
For problems 12 and 13, determine the missing numerator or denominator.
Exercise:
Problem: ((link]) 2 = 3;
Solution:
20
Exercise:
Problem: ({link]) ie = 20
Solution:
24
For problems 14-25, find each value.
Exercise:
Problem: ({link]) Ta ee
Solution:
3
20
Exercise:
mn: 3 9 3
Solution:
oo
Exercise:
Problem: ((Jink]) ,/ =
Solution:
Hor
Exercise:
Problem: ((link]) 4
Solution:
3
4
Exercise:
Problem: ({link]) a iy
Solution:
bole
Exercise:
Problem: ({link])
Solution:
=
30
Exercise:
Problem: ({link]) — 2
Solution:
=
11
4 .
7 aa
12
: 4
+24
Exercise:
Problem: ({link]) ys o / Los
Solution:
36 yl
op = 1
Exercise:
Problem: ({link]) Find 2 of 3.
Solution:
2
5
Exercise:
Problem: ([ link}) 4 = of what number is Tk
Solution:
1
4
Exercise:
Problem: ((link]) 1+ 2 of 2 is what number?
Solution:
Oy =>
$=]
on)
Exercise:
Problem: ({link]) What part of & is 4 ?
Solution:
Objectives
This module contains the learning objectives for the chapter "Addition and
Subtraction of Fractions" from Fundamentals of Mathematics by Denny
Burzynski and Wade Ellis, jr.
After completing this chapter, you should
Addition and Subtraction of Fractions with Like Denominators ({link]})
e be able to add and subtract fractions with like denominators
Addition and Subtraction of Fractions with Unlike Denominators
({link])
e be able to add and subtract fractions with unlike denominators
Addition and Subtraction of Mixed Numbers ({link])
e be able to add and subtract mixed numbers
Comparing Fractions ({link])
e understand ordering of numbers and be familiar with grouping
symbols
e be able to compare two or more fractions
Complex Fractions ({link])
¢ be able to distinguish between simple and complex fractions
e be able to convert a complex fraction to a simple fraction
Combinations of Operations with Fractions ({link])
e gain a further understanding of the order of operations
Addition and Subtraction of Fractions with Like Denominators
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to add and subtract fractions
with like denominators. By the end of the module students should be able to
add and subtract fractions with like denominators.
Section Overview
e Addition of Fraction With Like Denominators
e Subtraction of Fractions With Like Denominators
Addition of Fraction With Like Denominators
Let's examine the following diagram.
2 one-fifths and 1 one fifth is shaded.
It is shown in the shaded regions of the diagram that
(2 one-fifths) + (1 one-fifth) = (3 one-fifths)
That is,
2 15.
54 5 5
From this observation, we can suggest the following rule.
Method of Adding Fractions Having Like Denominators
To add two or more fractions that have the same denominators, add the
numerators and place the resulting sum over the common denominator.
Reduce, if necessary.
Sample Set A
Find the following sums.
Example:
3 sl 2. The denominators are the same. Add the numerators and place that
sum over 7.
se BID eee ne pomee Iulh
Ge ae
Example:
z =F 3. The denominators are the same. Add the numerators and place the
sum over 8. Reduce.
SE ase IGS he a
og =
8 Si 2
Example:
my se 2. The denominators are the same. Add the numerators and place the
sum over 9.
Example:
. ae 2. The denominators are the same. Add the numerators and place the
sum over 8.
tee pe li eee ee Pena a
She 8 8 2
Example:
To see what happens if we mistakenly add the denominators as well as the
numerators, let's add
1 1
— + —
ete
Peeing the numerators oe mistakenly adding the denominators produces
all r+ eae) lore eter Neen
~ 242 — _ = 5
This means that two 5 's is the same as one = 5 - Preposterous! We do not
add denominators.
Practice Set A
Find the following sums.
Exercise:
Problem: io + 40
Solution:
2
5
Exercise:
etl i
Problem: 7 ae
Solution:
1
2
Exercise:
Problem:
Solution:
Exercise:
AE | 1
Problem: more
Solution:
4
5
Exercise:
Problem:
Show why adding both the numerators and denominators is
preposterous by adding f and a and examining the result.
Solution:
Ee: Geen in gis ER 0 ee 2 Bra 3 sigs
4 + 4 = ga = gs = 7 SO two {’s= one 7 which is preposterous.
Subtraction of Fractions With Like Denominators
We can picture the concept of subtraction of fractions in much the same
way we pictured addition.
Mi fi] tke fifi yi]:
615 | 5 away 5 | 5 feb 5 | 5
5 - 8
(3 one-fifths) - (1 one-fifth)
(2 one-fifths)
From this observation, we can suggest the following rule for subtracting
fractions having like denominators:
Subtraction of Fractions with Like Denominators
To subtract two fractions that have like denominators, subtract the
numerators and place the resulting difference over the common
denominator. Reduce, if possible.
Sample Set B
Find the following differences.
Example:
4 — $ The denominators are the same. Subtract the numerators. Place the
difference over 5.
3 iL, Sail —. 2
cos 5 5
Example:
Seer?
ieee Te The denominators are the same. Subtract the numerators. Place the
difference over 6.
8 ea ee
6 6 6 6
Example:
— — = The denominators are the same. Subtract numerators and place
the difference over 9.
16 ee lp eet
9 9 9 9
Example:
To see what happens if we mistakenly subtract the denominators, let's
consider
fi 4 7=A 3
Whos plese eatin a e0)
We get division by zero, which is undefined. We do not subtract
denominators.
Practice Set B
Find the following differences.
Exercise:
gO. © BS
Problem: 3 i3
Solution:
2.
13
Exercise:
Problem: — — =>
Solution:
i
3
Exercise:
as | 1
Problem: eS
Solution:
0
Exercise:
aie, =~, ae
Problem: 0 0
Solution:
6
b)
Exercise:
Problem:
Show why subtracting both the numerators and the denominators is in
2
error by performing the subtraction 2 Ge
Solution:
5 9 _ -_ 3 . . .
3 — o = 9-9 = > which is undefined
Exercises
For the following problems, find the sums and differences. Be sure to
reduce.
Exercise:
8 9
Problem: amar
Solution:
5
8
Exercise:
ras 2
Problem: aor e
Exercise:
Problem: — + To
Solution:
1
Exercise:
Problem:
Exercise:
Problem:
Solution:
13
15
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
Exercise:
Problem:
Exercise:
Problem:
Solution:
0
Exercise:
12
16
Problem:
Exercise:
Problem:
Solution:
0
Exercise:
Problem:
Exercise:
Problem:
Solution:
=z
17
Exercise:
Problem:
Exercise:
Problem:
Solution:
15
8
Exercise:
Di = 32
23 23
ie
6 6
1 i a
atata
3 1 5
Ttiait i
HG). he nip. 3D
20 + 20 20
Oe Se Oe fe ak
8 +ets
Problem:
Exercise:
Problem:
Solution:
ai
2
Exercise:
Problem:
Exercise:
Problem: =~
Solution:
3
by)
Exercise:
Problem: —~ —
Exercise:
Problem:
Solution:
ne}
5
Exercise:
11
16
8 6
+ 15 a 15
eee
8 8
9 5
ag? AG.
1 9
— 29 F 20
B. 5.
10 10
me ee
5 5
Problem: 22 — 22 + 24
Exercise:
oO eh O35 he
Problem: > + 5 5
Solution:
10
Exercise:
Problem: + + — + +
18 18 18
Exercise:
e 6 ar 2 4A
Problem: 55 a BB
Solution:
9
af
The following rule for addition and subtraction of two fractions is
preposterous. Show why by performing the operations using the rule for the
following two problems.
Preposterous Rule
To add or subtract two fractions, simply add or subtract the numerators and
place this result over the sum or difference of the denominators.
Exercise:
oe EO
Problem: i0 i0
Exercise:
a: 8
Problem: Ip + 5
Solution:
ie = (using the preposterous rule)
Exercise:
Problem: Find the total length of the screw.
Exercise:
Problem:
Two months ago, a woman paid off oh of a loan. One month ago, she
paid off on of the total loan. This month she will again pay off = of
the total loan. At the end of the month, how much of her total loan will
she have paid off?
Solution:
13
24
Exercise:
Problem: Find the inside diameter of the pipe.
Exercises for Review
Exercise:
Problem: ([link]) Round 2,650 to the nearest hundred.
Solution:
2700
Exercise:
Problem:
({link]) Use the numbers 2, 4, and 8 to illustrate the associative
property of addition.
Exercise:
Problem: ({link]) Find the prime factors of 495.
Solution:
32-5-11
Exercise:
Problem: ({link]) Find the value of 3 : = : a
Exercise:
Problem: ({link}) . of what number is Le
Solution:
2
3
Addition and Subtraction of Fractions with Unlike Denominators
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This
module discusses how to add and subtract fractions with unlike denominators. By the end of the module
students should be able to add and subtract fractions with unlike denominators.
Section Overview
e A Basic Rule
e Addition and Subtraction of Fractions
A Basic Rule
There is a basic rule that must be followed when adding or subtracting fractions.
A Basic Rule
Fractions can only be added or subtracted conveniently if they have like denominators.
To see why this rule makes sense, let's consider the problem of adding a quarter and a dime.
1 quarter + 1 dime = 35 cents
Now,
1 quarter = —
. 10 same denominations
1 dime = a
— _35_
35,¢= i00
25, 10 _ 25410 _ 35
100 ' 100 ~ 100 ~~ 4100
In order to combine a quarter and a dime to produce 35¢, we convert them to quantities of the same
denomination.
Same denomination > same denominator
Addition and Subtraction of Fractions
Least Common Multiple (LCM) and Least Common Denominator (LCD)
In [link], we examined the least common multiple (LCM) of a collection of numbers. If these numbers
are used as denominators of fractions, we call the least common multiple, the least common
denominator (LCD).
Method of Adding or Subtracting Fractions with Unlike Denominators
To add or subtract fractions having unlike denominators, convert each fraction to an equivalent fraction
having as a denominator the least common denominator ( LCD) of the original denominators.
Sample Set A
Find the following sums and differences.
Example:
é ae 3. The denominators are not the same. Find the LCD of 6 and 4.
6=2-3
Anas
Write each of the original fractions as a new, equivalent fraction having the common denominator 12.
Sear sige
€? 4 = aD
To find a new numerator, we divide the original denominator into the LCD. Since the original
denominator is being multiplied by this quotient, we must multiply the original numerator by this
quotient.
12+-6=2
\oneLeD =2?-3=4-3=12
Multiply t by 2: 1-2=2.
original numerator
new numerator
12+-4=3
Multiply 3 by ? 3-3=9.
original numerator
new numerator
ee =o ara
= >= + > Now the denominators are the same.
= a Add the numerators and place the sum over the common denominator.
— i
12
Example:
$ + +. The denominators are not the same. Find the LCD of 2 and 3.
Re Di 23 =
Write each of the original fractions as a new, equivalent fraction having the common denominator 6.
aa 3 S65 5
To find a new numerator, we divide the original denominator into the LCD. Since the original
denominator is being multiplied by this quotient, we must multiply the original numerator by this
quotient.
6 + 2 = 3 Multiply the numerator 1 by 3.
6 + 2 = 3 Multiply the numerator 2 by 2.
Example:
2 a >: The denominators are not the same. Find the LCD of 9 and 12.
SiS Seae sae
LOD 27237 = Ao 36
2S 2G Se Dae) 8
5 —
Ge ay iia 0 ese
36+9 = 4 Multiply the numerator 5 by 4.
36+12 = 3 Multiply the numerator 5 by 3.
5 i ee eke 5.3
9 12 36 36
= A) — 1
— 36 36
20-15
36
ae
36
Example:
3 — 3 + ie The denominators are not the same. Find the LCD of 6, 8, and 16
= os
8 = 2-4=2-2.2=23 The LCD is 24 - 3 = 48
LG A058 = 00 AO 27
ie gl ees
6 sti¢e 8 wT @
48~6 = 8 Multiply the numerator 5 by 8
48~8 = 6 Multiply the numerator 1 by 6
48+16 = 3 Multiply the numerator 7 by 3
Oke s i ge a alas 73
Ge ceae tae ea == Sas 43 + 48
Ai oul
Sig ee TAR
40-6421
48
= Sky nes
= ese
Practice Set A
Find the following sums and differences.
Exercise:
Problem: ~ + —
Solution:
5
6
Exercise:
1 3
Problem: aan J
Solution:
ais
14
Exercise:
Problem:
Solution:
aco
40
Exercise:
215 7 1 3
Problem: ig 15 ri
Solution:
ao
16
Exercise:
Problem: => — ->
Solution:
1.
96
Exercises
Exercise:
Problem:
A most basic rule of arithmetic states that two fractions may be added or subtracted conveniently
only if they have .
Solution:
The same denominator
For the following problems, find the sums and differences.
Exercise:
Problem:
Exercise:
Problem:
Solution:
5
8
Exercise:
Problem:
Exercise:
Problem:
Solution:
3L
24
Exercise:
Problem:
Exercise:
Problem:
Solution:
ap
28
Exercise:
Problem:
Exercise:
Problem:
Solution:
19
36
Exercise:
| co
+
colon
+
wl
eo|bo
Problem:
Exercise:
Problem:
Solution:
19
39
Exercise:
Problem:
Exercise:
Problem:
Solution:
29
60
Exercise:
Problem:
Exercise:
Problem:
Solution:
8
81
Exercise:
Problem: —
Exercise:
Problem:
Solution:
Big
65
Exercise:
Problem:
CMa
15 10
ues 5, 2B
13 39
15 2
12 5
1 5
ip + aa
18 i.
88 4
ose
9 81
19 , 5
10 + 2
2
26 10
Loe
28 45
Exercise:
Problem:
Solution:
2
63
Exercise:
Problem:
Exercise:
Problem:
Solution:
7
16
Exercise:
Problem:
Exercise:
Problem:
Solution:
47
18
Exercise:
Problem:
Exercise:
Problem:
Solution:
103
30
Exercise:
Problem:
Exercise:
16
17
10
Problem:
Solution:
Exercise:
Problem:
Exercise:
Problem:
Solution:
Exercise:
Problem:
Exercise:
Problem:
Solution:
37
72
Exercise:
Problem:
Exercise:
Problem:
Solution:
Exercise:
Problem:
Exercise:
Problem:
3 3 5
4 22 1 24
25 Ty 5
48 — 33 + 24
27 =, 47 _ 119
40 + 48 126
BA oj GS on ST
44. 99 ~ 175
fey he ae AL
12 a 18 a 24
ee eee
og a5
uae ta Heo
18 36 + 9
Hy oe De elk
14.36 32
Solution:
1,465
2,016
Exercise:
Problem:
Exercise:
Problem:
Solution:
65_
204
Exercise:
Problem:
Exercise:
Problem:
Solution:
i
5
Exercise:
Problem:
Exercise:
Problem:
Solution:
Exercise:
Problem:
Exercise:
Problem:
Solution:
21 12 15
33 + a + 5
he ge Ds, Gn ale
51 + 34 + 68
8 16, 19
7 14 a 21
7, 3 _ 34
15 a 10 60
14 3 6 €
15 10 25 20
11 ee eee ve 25
6 12 ' 30 18
V0 122 5 1
9 ' 21 18 45
7 | 28 51 0
26 ' 65 104
109
520
Exercise:
Problem:
A morning trip from San Francisco to Los Angeles took 2 hours. The return trip took i hours.
6
How much longer did the morning trip take?
Exercise:
Problem:
At the beginning of the week, Starlight Publishing Company's stock was selling for aa dollars per
share. At the end of the week, analysts had noted that the stock had gone up ub dollars per share.
What was the price of the stock, per share, at the end of the week?
Solution:
137 1
$ a or$17 e
Exercise:
Problem:
2 cups of
orange juice, 2 cups of sugar, 6 cups of water, and 8 cups of carbonated non-cola soft drink. How
many cups of ingredients will be in the final mixture?
A recipe for fruit punch calls for B cups of pineapple juice, - cup of lemon juice,
Exercise:
Problem:
The side of a particular type of box measures 83 inches in length. Is it possible to place three such
boxes next to each other on a shelf that is 26+ inches in length? Why or why not?
Solution:
No; 3 boxes add up to 26+ , which is larger than 25.
Exercise:
Problem:
Four resistors, ° ohm, i ohm, 2 ohm, and tL ohm, are connected in series in an electrical circuit.
What is the total resistance in the circuit due to these resistors? ("In series" implies addition.)
Exercise:
Problem:
A copper pipe has an inside diameter of 2 = inches and an outside diameter of 22 inches. How
thick is the pipe?
Solution:
No pipe at all; inside diameter is greater than outside diameter
Exercise:
Problem:
The probability of an event was originally thought to be =. Additional information decreased the
probability by a. What is the updated probability?
Exercises for Review
Exercise:
Problem: ([link]) Find the difference between 867 and 418.
Solution:
449
Exercise:
Problem: ([link]) Is 81,147 divisible by 3?
Exercise:
Problem: ([link]) Find the LCM of 11, 15, and 20.
Solution:
660
Exercise:
Problem: ([link]) Find + of 42.
Exercise:
Problem: ((Link]) Find the value of — — 4 + +.
Solution:
=e
15
Addition and Subtraction of Mixed Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade
Ellis, Jr. This module discusses how to add and subtract mixed numbers. By the end
of the module students should be able to add and subtract mixed numbers.
Section Overview
e The Method of Converting to Improper Fractions
To add or subtract mixed numbers, convert each mixed number to an improper
fraction, then add or subtract the resulting improper fractions.
Sample Set A
Find the following sums and differences.
Example:
82 oe 5+. Convert each mixed number to an improper fraction.
gd — 58+3 _ 40+3 _ 43
5
7 gate Pen ae
43 4 21 The LCD = 20.
Ce ee Ser 21-5
5 a 4 a 20 ae 20
eA, 105
20 os 20
172+105
20
am Convert this improper fraction to a mixed number.
5
oo = = =| Now add the improper fractions Sauls.
17
3 - cs il
Example:
a = a Convert the mixed number to an improper fraction.
gil _ 3:8+1 _ 2441 _ 25
8 8 8 8
= = = The ice Di= 4
PAS apane Cip prmney ir eat |
8 Ga 4 24
Be el
24 24
75—20
24
eerie
OR
7
ee uM 7
Thus, 33 : =253
Practice Set A
Convert his improper fraction to a mixed number.
Find the following sums and differences.
Exercise:
Problem: 1 + 3+
Solution:
i
45
Exercise:
-1n3 1
Problem: 107 _ 25>
Solution:
1
87
Exercise:
Problem: 2— + 5
oo|~a
Al
Solution:
1
83
Exercise:
~Q3
Problem: B= —~ Fo
Solution:
3
8 ig
Exercise:
Problem: 16 + 22
Solution:
Exercises
For the following problems, perform each indicated operation.
Exercise:
Problem: 3~ + 42
Solution:
1
(ey
Exercise:
Problem: 5+ + 65
Exercise:
: 5 1
Problem: 1035 + 255
Solution:
1
124
Exercise:
Problem: 154 — 112
Exercise:
Problem:
Solution:
Exercise:
Problem:
Exercise:
Problem:
Solution:
3
oo|~NI
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
wb
Exercise:
Problem
Exercise:
Problem
: 2
65> 114
217-84
Solution:
11
Say
Exercise:
Problem: 54+ + 2+
3 4
Exercise:
Problem: 62 = 1}
Solution:
20
4sT
Exercise:
Problem: 82 + 44
Exercise:
Problem: 15 ae 122
Solution:
17
1354
Exercise:
Problem: 3+ + 13 91
Exercise:
Problem: 43 — 33 +12
Solution:
12
Exercise:
Problem: 335 + 45 +14
Exercise:
Sele ig s8) a2
Problem: 55 + 855 — 5=
Solution:
17
7 30
Exercise:
Problem: 7+ + 82 — 2+
Exercise:
Problem: 1922 + 42% — 3 +122
Solution:
25
Aaa
Exercise:
1 3 3
Problem: 7; + 47 + 10; —9
Exercise:
Problem: 11 — - + 105 = - = 5s +645
Solution:
1
21 =
Exercise:
Problem: = + 2+ +114 —-—
2 6 3 6
Exercise:
a7 9 12 _ 21 19
Problem: 1 era ia at
Solution:
21
9 35
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
20 sz
Exercise:
Problem:
Exercise:
Problem:
Solution:
7
Vay
Exercise:
Problem:
Exercise:
Problem:
Solution:
13
5 132
Exercise:
Problem:
222 — 162
15¢ +43
17 9
11 2
67 +3
9 ce
856 — 9
2 1
oT — i
15 33
Exercise:
489 21
Problem: Le 5G
Solution:
AT
1535
Exercise:
Ree a Gl 13
Problem: 1157 _ las
Exercise:
627 5 1
Problem: oar - Say + Lee
Solution:
1
37
Exercise:
: 1 1 1
Problem: 1645 = 1635 + 144
Exercise:
Problem:
A man pours 22 gallons of paint from a bucket into a tray. After he finishes
pouring, there are 1; gallons of paint left in his bucket. How much paint did the
man pour into the tray?
Note:Think about the wording.
Solution:
22 gallons
Exercise:
Problem:
A particular computer stock opened at 37 3 and closed at 384 . What was the
net gain for this stock?
Exercise:
Problem:
A particular diet program claims that 43. pounds can be lost the first month,
34 pounds can be lost the second month, and 15 pounds can be lost the third
month. How many pounds does this diet program claim a person can lose over a
3-month period?
Solution:
8 = pounds
Exercise:
Problem:
If a person who weighs 145 3 pounds goes on the diet program described in the
problem above, how much would he weigh at the end of 3 months?
Exercise:
Problem:
If the diet program described in the problem above makes the additional claim
that from the fourth month on, a person will lose 12 pounds a month, how much
will a person who begins the program weighing 208+ pounds weight after 8
months?
Solution:
3
194-7 pounds
Exercises for Review
Exercise:
Problem: ([link]) Use exponents to write 4 - 4 - 4.
Exercise:
Problem: ({link]) Find the greatest common factor of 14 and 20.
Solution:
2
Exercise:
Problem: ({link]) Convert 2 to a mixed number.
Exercise:
Problem: (({link]) Find the sum. 5; =f 7 =f 2.
Solution:
co|NI
Exercise:
Problem: (({link]) Find the difference. or — .
Comparing Fractions
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to compare fractions. By the
end of the module students should be able to understand ordering of
numbers and be familiar with grouping symbols and compare two or more
fractions.
Section Overview
e Order and the Inequality Symbols
e¢ Comparing Fractions
Order and the Inequality Symbols
Our number system is called an ordered number system because the
numbers in the system can be placed in order from smaller to larger. This is
easily seen on the number line.
0 12 3 4 5 6 7 8 9 10
-———- smaller larger-———~
On the number line, a number that appears to the right of another number is
larger than that other number. For example, 5 is greater than 2 because 5 is
located to the right of 2 on the number line. We may also say that 2 is less
than 5.
To make the inequality phrases "greater than" and "less than" more brief,
mathematicians represent them with the symbols > and <, respectively.
Symbols for Greater Than > and Less Than <
> represents the phrase "greater than."
< represents the phrase "less than."
5 > 2 represents "5 is greater than 2."
2 <5 represents "2 is less than 5."
Comparing Fractions
Recall that the fraction < indicates that we have 4 of 5 parts of some whole
quantity, and the fraction 2 indicates that we have 3 of 5 parts. Since 4 of 5
parts is more than 3 of 5 parts, 4 is greater than 3; that is,
a
Bo 2
We have just observed that when two fractions have the same denominator,
we can determine which is larger by comparing the numerators.
Comparing Fractions
If two fractions have the same denominators, the fraction with the larger nu-
merator is the larger fraction.
Thus, to compare the sizes of two or more fractions, we need only convert
each of them to equivalent fractions that have a common denominator. We
then compare the numerators. It is convenient if the common denominator
is the LCD. The fraction with the larger numerator is the larger fraction.
Sample Set A
Example:
Compare - and a
Convert each fraction to an equivalent fraction with the LCD as the
denominator. Find the LCD.
a 9
Phek@D 34-5. —9- 5 —A5
15 4)
sere aU
9 45 45
Since 40 < 42,
On eau
ri ce et,
Thus 9 —< 15°
Example:
Write ee and = in order from smallest to largest.
Convert each fraction to an equivalent fraction with the LCD as the
denominator.
Find the LCD.
Chena
lO 25 he PCD 2 32-5. — 30)
1 Stra en
5 55 25
6 30 30
Sine Seat
15 30
21 < 25 < 26
(lege Aaa
ea ere
16r -G a5
Writing these numbers in order from smallest to largest, we get =. 2 =
Example:
Compare 84 and 63.
To compare mixed numbers that have different whole number parts, we
need only compare whole number parts. Since 6 < 8,
3 6
62 <8*
Example:
Compare 4 2and4 ay
To compare mixed numbers that have the same whole number parts, we
need only compare fractional parts.
8 = 23 3
5 ThekCD = 2) 3 — sero 4
1b emer rs
ciN ee pale ei
8 24 24
Pe ie eee 1 ree eS
{Ons mated
Since 14 < 15
14 15
mS 2
i
7 5
Hence, Ls ae aa
Practice Set A
Exercise:
Problem: Compare * and .
Solution:
3 4
ae
Exercise:
Problem: Compare ~ =o and 3.
Solution:
13
=s
tS
10
Exercise:
Problem: Write +2 2. P ae , and 2 in order from smallest to largest.
Solution:
13° 33) 17
16’ 40 ’ 20
Exercise:
Problem: Compare 114 and 92.
Solution:
92 < 115
Exercise:
Problem: Compare 14 and liz.
Solution:
9 11
Exercises
Arrange each collection of numbers in order from smallest to largest.
Exercise:
olen
Problem: 3,
Solution:
b)
5 8
Exercise:
Problem: a
~I|b
Exercise:
oo
Problem: +,
Solution:
3 5
a> 6
Exercise:
e 7 —-
Problem: =, =5
Exercise:
Problem:
Solution:
3 2
BS 8
Exercise:
Problem:
Exercise:
Problem:
Solution:
4
ea
Exercise:
Problem:
Exercise:
3 2
8? 5
i 5
a8
te 3B
2E%
3
5
3 2
4? 3
Ne
ro ee ed
Problem: =, oe a
Solution:
3 7 5
fog ee
Exercise:
Mica, «, GAO 0 cele
Problem: CATR
Exercise:
Pee: Se ae
Problem: =7, 7, {
Solution:
3 2 3
1
Exercise:
6 EE es 18,
Problem: =>, 4, 73
Exercise:
.r3 64
Problem: 5= , 5>
Solution:
4 3
5 7 25 5
Exercise:
, 3 1
Problem: 11 55, 1135
Exercise:
-Qg2 o4
Problem: 9 ev 9
Solution:
2 4
ard < a
Exercise:
AG a: 5
Problem: ere 8s
Exercise:
ee
Problem: 157,
1
239
Solution:
9 1
Exercise:
Problem: 2032, 2033
Exercise:
Problem: 22, 24
Solution:
2 3
Exercise:
Problem: 5
9
13? 920
Exercises for Review
Exercise:
Problem: ({link]) Round 267,006,428 to the nearest ten million.
Solution:
270,000,000
Exercise:
Problem: (({link]) Is the number 82,644 divisible by 2? by 3? by 4?
Exercise:
Problem: ({link]) Convert 32 to an improper fraction.
Solution:
23
7
Exercise:
Problem: ((Link]) Find the value of 2 + 35 — 2
Exercise:
Problem: ((link]) Find the value of gs fs 55.
Solution:
5 109
13 yor 3.
Complex Fractions
This module is from Fundamentals of Mathematics by Denny Burzynski and
Wade Ellis, Jr. This module discusses complex fractions. By the end of the
module students should be able to distinguish between simple and complex
fractions and convert a complex fraction to a simple fraction.
Section Overview
e Simple Fractions and Complex Fractions
e Converting Complex Fractions to Simple Fractions
Simple Fractions and Complex Fractions
Simple Fraction
A simple fraction is any fraction in which the numerator is any whole
number and the denominator is any nonzero whole number. Some examples
are the following:
1 4 _763
2° 3? 1,000
Complex Fraction
A complex fraction is any fraction in which the numerator and/or the
denominator is a fraction; it is a fraction of fractions. Some examples of
complex fractions are the following:
Converting Complex Fractions to Simple Fractions
The goal here is to convert a complex fraction to a simple fraction. We can
do so by employing the methods of adding, subtracting, multiplying, and
dividing fractions. Recall from [link] that a fraction bar serves as a grouping
symbol separating the fractional quantity into two individual groups. We
proceed in simplifying a complex fraction to a simple fraction by simplifying
the numerator and the denominator of the complex fraction separately. We
will simplify the numerator and denominator completely before removing the
fraction bar by dividing. This technique is illustrated in problems 3, 4, 5, and
6 of [link].
Sample Set A
Convert each of the following complex fractions to a simple fraction.
Convert this complex fraction to a simple fraction by performing the
indicated division.
ss
ae Sea The divisor is; . Invert =2-and multiply.
1 2
a eee eee
a v4 DS del
1 5
Example
4
ar Write 6 as? and divide.
m 4 2a
a 9° 1
1
2
my Sag yey Cleans
— 9 ¥ ~ 93 ~ 27
3
Example:
3
aa Simplify the numerator.
4-543 20+3 23 te
4 a= 4 pany 4 7 ae
6° = ae ae Write 46 as ine
ie eased.
ne a
il
ill
ey Pei otter eal eae lt
_ 4 Me 2 8
2
Example:
cee es 243 5
PSB ees SOR ee ee Se Be te oo
il i) oe 12 137 — 12413 20
a +34 ator i es es
1 8)
Bunge as oe ae» cae
ogee yy pe | I aS
il 5
Example:
5 4.6+5 29
Ad sod eet panna Oe ee ea
maiee Sse ay 120) Gas aG 3
3 33 a
1
ey ae)
= 7 = oo
2
Example:
33 11-10+3 110+3 iil}
11+ a9 te — i = WwW. =, JS 6 4
44 side =) 0s a ee = HN) is
5) 5 5 5)
Practice Set A
Convert each of the following complex fractions to a simple fraction.
Exercise:
Problem:
Solution:
lon
Exercise:
Problem:
Solution:
=
40
Exercise:
Problem:
Solution:
3
7)
Exercise:
Problem:
Solution:
10
57
Exercise:
Problem:
eiffel
Solution:
13;
2 22
Exercise:
Probl 16-102
roblem: — 5-77
Solution:
5
lay
Exercises
Simplify each fraction.
Exercise:
Problem:
Bofors
Solution:
1
Exercise:
Problem:
Exercise:
Problem:
Solution:
3.
5
Exercise:
8
Problem: —
cr
Exercise:
Problem:
11+4
Solution:
B)
9
Exercise:
2+5
Problem: i
aed
Exercise:
Problem: = —-
Solution:
H)
2
Exercise:
9+
Problem: 5
Lea
Exercise:
10
13
Problem: 42
39
31
2
Exercise:
1
Problem: 3 te
See
Problem: £—
Solution:
7
Se
Problem: 10 Te
Problem: 16' 3
1
ei
Problem: =~
Exercise:
Problem: =
Solution:
alm
Exercise:
Problem:
Exercise:
Problem: ———
Solution:
52
81
Exercise:
Problem:
Exercise:
Problem:
Solution:
16
21
Exercise:
Problem:
Exercise:
Problem:
Solution:
686
101
Exercise:
54345
Problem: —*—2-
8g -45
2 11
11 3 547
Exercise:
53434
2241555
Problem: ae ae a
95-44
$2797
Solution:
i
3
Exercises for Review
Exercise:
Problem: ({link]) Find the prime factorization of 882.
Exercise:
Problem: ({link]) Convert o to a mixed number.
Solution:
6
87
Exercise:
Problem: ({link]) Reduce a5 to lowest terms.
Exercise:
Problem: ({link]) Find the value of 62 — 42.
Solution:
13. 37
log OF 34
Exercise:
Problem: ({link]) Arrange from smallest to largest:
Combinations of Operations with Fractions
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses combinations of operations with
fractions. By the end of the module students should gain a further
understanding of the order of operations.
Section Overview
e The Order of Operations
The Order of Operations
To determine the value of a quantity such as
aa
oe ap
where we have a combination of operations (more than one operation
occurs), we must use the accepted order of operations.
The Order of Operations:
1. In the order (2), (3), (4) described below, perform all operations inside
grouping symbols: ( ), [ ], (), —. Work from the innermost set to the
outermost set.
2. Perform exponential and root operations.
3. Perform all multiplications and divisions moving left to right.
4. Perform all additions and subtractions moving left to right.
Sample Set A
Determine the value of each of the following quantities.
Example:
pees eas
a. Multiply first.
b. Now perform this addition. Find the LCD.
A= 2?
The CD01).
12 = 22.3
il 1 ilo ie* oe. 3} 1
ais pase Po he = ee ep)
wet eoeely eae ae See ade
Oe ee bk ae
1 5 2 eel
Thus, 7+ 3° 5 = 3
a. Operate within the parentheses first, (2 = +).
=P
— 3
v2 pThe LCD = 27.3? = 4.9 =36.
Now we have
3 9 (11
3 + a (ze)
b. Perform the multiplication.
4 4
c. Now perform the addition. The LCD=80.
3 Ie S56 gay) eras one ds ers
Gana iit = 060m i sO = 28), Geen Sed == 0
Example:
15 4 1 i
— gag (2 — 155) (35 +25)
a. Work within each set of parentheses individually.
4 _ 11544 9 19
7s Can 2 (t= Ss 15
0 et ee BO Oe en eal
15 15 vey tomes ty
ab [ee es We ee
Ore, tegretol ag
16 My
Ba ye
16-8 17-5
40 eh 40
128 4. 85
40. +t 40
128485
40
213
40
Now we have
~ as (G5) ao")
b. Now multiply.
af 1
Sie | cut a iil
pr aa so 40 21-40 80
2 ‘I
c. Now subtract.
oy ee OE SCE a ee Se es
80 — 80 80 — 80 80 sae csr 680
69
or (Gre
Thus, 8 — ¢x(2-14)(3¢ +24) =78
Example:
ey eae
4 9 12
3
a. Square {.
i 1
Oe ere ce wee eae ate Gee
Sie 2 il ip 2 12
2 1
c. Now perform the subtraction.
Glee hes sla trey (costae he
ae) ian to 12 12
Example: _
7 OS og 1 1
Eg Ee ok dy
yl ee yy eee ee ef ee en ee
2 3 2 3 nes
ols eee Oe
ene Ca cole oe 8
b. Now simplify the square root.
25 5 25
- $ (since($) = a)
Now we have
7 5b.
&
6
c. Perform the division.
cf
Tie Wain Ee mene A Ra I rans Paes
SR eye ce = hadi ae Sauna
1
d. Now perform the addition.
re Bee ese pies é =
D257, 586 4 40
56 ae 56° ~—SoiBG ae 56
— 161440 _ 201 aa)
= 5 OT Gg (OF 3 5G
7 yea ca teen 33
Thus, 22 + 4/28 + (24-14) =3=
Practice Set A
Find the value of each of the following quantities.
Exercise:
Problem: — - = — =
Solution:
0
Exercise:
6 21. 9
Problem: 2 - = ~ —
7 40 ° 10
Solution:
35
35 5a
7 or 94
Exercise:
Problem: 855 — 2(45 —
Solution:
oT 1
30 «OF 139
Exercise:
Problem: 22 — 2° (+ =
18 30
Solution:
ve
9
Exercise:
Problem: (<5 — 15) + (1
Solution:
3
3)
6
27
Exercise:
Problem: =———
Solution:
3
11
Exercise:
Problem: en a “ :
Solution:
15
64
Exercise:
322 1
Problem: ae 25 —-
Solution:
it
10
Exercises
Find each value.
Exercise:
oo a le
Problem: 3 gs
Solution:
b)
4
Exercise:
Problem:
Exercise:
Problem:
Solution:
2
47
Exercise:
Problem:
Exercise:
Problem:
Solution:
2
15
Exercise:
Problem:
Exercise:
Problem:
Solution:
woe
9 5
2 5
27+ %
Be ae, Os
16 14
4 8
25 15
2 i
ae ers
3 (3
ot
Exercise:
Problem: 32 . (8 — 2)+2-(2+
Exercise:
Problem:
“|
Solution:
3
O77
Exercise:
Problem:
Exercise:
foro)
2
Problem: (+) +
Solution:
3
8
Exercise:
Problem: . = #4
Exercise:
iin BO ey Se
Problem: aT tas
Solution:
20
27
Exercise:
Problem: ,/ an — /2
Exercise:
ee eee eee ee
Problem: /2 ; one
Solution:
0
Exercise:
Problem: (3)° — =
Exercise:
Problem: ( = )
Solution:
2
5
Exercise:
Problem: (/ >
Exercise:
Problem: (
ae fe es
/2++1
Solution:
1
Exercise:
: 3 1 i eee Ae |
Problem: ieee is ie
Exercise:
Problem: ree ey: ae
Solution:
gon
72
Exercise:
3 5
Problem: \/ 4 . fs gt
Exercise:
cle ae ( 3
Problem: £5 (z)
67 1 9
atta) Ma)
Solution:
252
19
Exercise:
: 16,1
Problem: jz +1.-6
Exercise:
oo] A
_.| [si 3
Problem: 256 32 -1
Solution:
165
256
Exercises for Review
Exercise:
Problem:
({link]) True or false: Our number system, the Hindu-Arabic number
system, is a positional number system with base ten.
Exercise:
Problem:
({link]) The fact that 1 times any whole number = that particular whole
number illustrates which property of multiplication?
Solution:
multiplicative identity
Exercise:
Problem: ({link]) Convert 82 to an improper fraction.
Exercise:
Problem: ({link]) Find the sum. 3 ai < + a
Solution:
241 1
120 Of 279
Exercise:
Problem: ({link]) Simplify
6+5
6-+
Summary of Key Concepts
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module reviews the key concepts from the chapter
"Addition and Subtraction of Fractions, Comparing Fractions, and Complex
Fractions."
Summary of Key Concepts
Addition and Subtraction of Fractions with Like Denominators ([{link])
To add or subtract two fractions that have the same denominators, add or
subtract the numerators and place the resulting sum or difference over the
common denominator. Reduce, if necessary. Do not add or subtract the
denominators.
At Oe de ee Bo 8
gt gs @ 8 4
Basic Rule for Adding and Subtracting Fractions ((link])
Fractions can be added or subtracted conveniently only if they have like
denominators.
Addition and Subtraction of Fractions with Unlike Denominators
({link])
To add or subtract fractions having unlike denominators, convert each
fraction to an equivalent fraction having as denominator the LCD of the
original denominators.
Addition and Subtraction of Mixed Numbers ([{link])
1. To add or subtract mixed numbers, convert each mixed number to an
improper fraction, then add or subtract the fractions.
Ordered Number System ([link])
Our number system is ordered because the numbers in the system can be
placed in order from smaller to larger.
Inequality Symbols ([link])
> represents the phrase "greater than."
< represents the phrase "less than."
Comparing Fractions ((link])
If two fractions have the same denominators, the fraction with the larger
numerator is the larger fraction.
5» 3
37 8
Simple Fractions ((link])
A simple fraction is any fraction in which the numerator is any whole
number and the denominator is any nonzero whole number.
Complex Fractions ((link])
A complex fraction is any fraction in which the numerator and/or the
denominator is a fraction.
Complex fractions can be converted to simple fractions by employing the
methods of adding, subtracting, multiplying, and dividing fractions.
Exercise Supplement
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is an exercise supplement for the chapter
"Addition and Subtraction of Fractions, Comparing Fractions, and Complex
Fractions" and contains many exercise problems. Odd problems are
accompanied by solutions.
Exercise Supplement
Addition and Subtractions of Fractions with Like and Unlike
Denominators, and Addition and Subtraction of Mixed Numbers
({link}, [link], [link])
For problems 1-53, perform each indicated operation and write the result in
simplest form.
Exercise:
2s 5
Problem: a0 =
Solution:
Exercise:
Problem: —-
Exercise:
will 3
Problem: gts
Solution:
1
2
Exercise:
5 1 5
Problem: 7 + 4g 7 oT
Exercise:
. 5 1 5
Problem: Qo 30 oT
Solution:
i ey Pe
nlp
Exercise:
<2 1
Problem: 5 Te
Exercise:
1 1 1
Problem: oe eT oe
Solution:
b)
8
Exercise:
Problem: ie io
Exercise:
Pe on ee
Problem: 7 ae 3
Solution:
13
21
Exercise:
° 1 1
Problem: ae + <=
Exercise:
Problem: 345 A 8
Solution:
455
Exercise:
Problem: 55 Ere 31
Exercise:
2 1
Problem: 16> + 87
Solution:
3
850
Exercise:
ul 4
Problem: 15 + 27>
Exercise:
Problem: 12 +0
Solution:
1
00 |eo
Exercise:
Problem: 35 +4
Exercise:
Problem: 182 +6
Solution:
2
24 =
Exercise:
Problem: 14 + 52
Exercise:
Problem: ~- = ie ~
Solution:
(in ae
0 ea 3 12
Exercise:
oy eee
Problem: i6 5
Exercise:
ie. eee
Problem: ii 55
Solution:
13
22
Exercise:
Problem: 6-2 — 12
Exercise:
Problem: 54 92 9
Solution:
1124
Exercise:
Problem: 8 — 42 = 3
Exercise:
s At
Problem: => + > — ig
Solution:
139
144
Exercise:
Problem: 72 — 52 —]
Exercise:
Problem: 162 — 84 — 3-2
Solution:
10
Exercise:
Problem: 4<. ay | ae
Exercise:
Problem: 4 ey | ee
Solution:
Ay
8
Exercise:
Problem: 8 — 2+
Exercise:
Problem: 4 — 3—-
Solution:
A
16
Exercise:
Problem: 63 +4
Exercise:
Problem: 114 — 3
Solution:
2
Sar
Exercise:
By 5
Problem: 21; ac
Exercise:
Pras: ae ae: See
Problem: + + 73 - =
Solution:
1
Exercise:
Problem: +5 + 42 +24
Exercise:
° 3 2 °
Problem: 155 + 23 +
co| es
Solution:
a
7 10
Exercise:
Problem: 82 —1— . 2
Exercise:
Problem: 23 +3
Solution:
b)
9
Exercise:
Problem: 152 + 50 — 35
Complex Fractions and Combinations of Operations with Fractions
({link], [Link])
Exercise:
pt
ale
Problem:
bo}
pan
w
bo
Solution:
6
4
Exercise:
Problem:
Exercise:
Problem:
Solution:
bole
a9
or 15
Exercise:
Problem:
aye |e
ojnyou
ol
pa
Exercise:
9
lig
it
23
Problem:
Solution:
15
28
Exercise:
8 A
Problem: oa
Exercise:
1
975
6
Problem:
Solution:
163 5
19g OF Lipg
Exercise:
3 =
Problem:
Exercise:
Problem: ——~-
Solution:
66 AL
13 OF 953
Exercise:
Problem:
Exercise:
Problem:
Solution:
xe
40
Exercise:
Problem:
Exercise:
Problem:
Solution:
255 71
ig OF Lagq
Comparing Fractions ((link])
For problems 54-65, place each collection in order from smallest to largest.
Exercise:
et,
Problem: =~, =;
Exercise:
Problem:
Solution:
3 tL
327° 8
Exercise:
Problem: —
Exercise:
Problem: —
Solution:
3 5.
10° 6
Exercise:
Problem:
Exercise:
Problem:
Solution:
3 8
S33
’
Exercise:
Problem:
Exercise:
Problem:
Solution:
5. UA
9° 7
Exercise:
Problem:
Exercise:
Problem:
Solution:
2: ee ee
19°97 5
Exercise:
Problem: =~
Exercise:
Problem:
Solution:
13
Proficiency Exam
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is a proficiency exam to the chapter
"Addition and Subtraction of Fractions, Comparing Fractions, and Complex
Fractions." Each problem is accompanied with a reference link pointing
back to the module that discusses the type of problem demonstrated in the
question. The problems in this exam are accompanied by solutions.
Proficiency Exam
For problems 1-12, perform each indicated operation and write the result in
simplest form.
Exercise:
Problem: ((link]) 4 +
Solution:
se
16
Exercise:
Problem: ((link|) 22 al 54
Solution:
75
Exercise:
Problem: (({link]) — 9 Oe
Solution:
1
Exercise:
Problem: ((link]) TLE + sce
Solution:
8
11
Exercise:
Problem: ((link]) 6+ - 150 — (3 7
Solution:
8
Exercise:
Problem
: ([link]) 5 — 24
Solution:
weg
27
Exercise:
Problem
: ({link]) 2
Solution:
49 17
32 Ol 30
Exercise:
143
Problem: ({Link]) —
1
Solution:
ate
15
Exercise:
Problem
: ([link)) 495 +13
Solution:
v4
376
Exercise:
Problem
: (Link) 3 - (4g —
Solution:
-s
32
Exercise:
Problem
: ((link]) 4+ 24
Solution:
1 19
65 or =
Exercise:
Problem:
(Hlink]) 87 —5
Solution:
3
NI|oo
—2
3
8
)
5
24
For problems 13-15, specify the fractions that are equivalent.
Exercise:
Problem: ((link]) =, +2
Solution:
equivalent
Exercise:
Problem: ((link]) >, =
Solution:
not equivalent
Exercise:
Problem: ((link]) >, 43
Solution:
equivalent
For problems 16-20, place each collection of fractions in order from
smallest to largest.
Exercise:
Problem: ({link]) S. £
Solution:
6
7
<o|00
’
Exercise:
Problem: ((link]) 2 t
Solution:
5 67
8° 9
Exercise:
Problem: ({link]) 11->
Solution:
ie Tat
Exercise:
5
Le
16’
5
15
Problem: ((link]) +>, +4, ¢
Solution:
Exercise:
Problem: ((link]) = _ ; =, 3
Solution:
9 19 5
16? 32° 8
B)
Objectives
This module contains the learning objectives for the chapter "Decimals"
from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, jr.
After completing this chapter, you should
Reading and Writing Decimals ({link])
e understand the meaning of digits occurring to the right of the ones
position
¢ be familiar with the meaning of decimal fractions
e be able to read and write a decimal fraction
Converting a Decimal to a Fraction ({link]})
e be able to convert an ordinary decimal and a complex decimal to a
fraction
Rounding Decimals ({link])
¢ be able to round a decimal number to a specified position
Addition and Subtraction of Decimals ({link])
¢ understand the method used for adding and subtracting decimals
e be able to add and subtract decimals
e be able to use the calculator to add and subtract decimals
Multiplication of Decimals ({link])
e understand the method used for multiplying decimals
e be able to multiply decimals
e be able to simplify a multiplication of a decimal by a power of 10
¢ understand how to use the word "of" in multiplication
Division of Decimals ({link])
¢ understand the method used for dividing decimals
e be able to divide a decimal number by a nonzero whole number and by
another, nonzero, decimal number
¢ be able to simplify a division of a decimal by a power of 10
Nonterminating Divisions ({link])
e understand the meaning of a nonterminating division
e be able to recognize a nonterminating number by its notation
Converting a Fraction to a Decimal ({link])
e be able to convert a fraction to a decimal
Combinations of Operations with Decimals and Fractions ({link])
e be able to combine operations with decimals
Reading and Writing Decimals
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to read and write decimals.
By the end of the module students should understand the meaning of digits
occurring to the right of the ones position, be familiar with the meaning of
decimal fractions and be able to read and write a decimal fraction.
Section Overview
Digits to the Right of the Ones Position
e Decimal Fractions
e Reading Decimal Fractions
e Writing Decimal Fractions
Digits to the Right of the Ones Position
We began our study of arithmetic ([link]) by noting that our number system
is called a positional number system with base ten. We also noted that each
position has a particular value. We observed that each position has ten times
the value of the position to its right.
| of 4 3 @
ra] = 2
s cE 86 E 5 :
S ts 68 8 cb & 6
LS) LN Ld Lr
10 X 100,000 10 X 10,000 10 X 1,000 10 100 10X10 10X1 1
This means that each position has TH the value of the position to its left.
as
10
as
+ x 100 4 x10
10
1 1
1,000,000 X 1,000,000 To X 100,000 — x 10,000 To X 1,000 10 10
Thus, a digit written to the right of the units position must have a value of
aT of 1. Recalling that the word "of" translates to multiplication (-), we can
see that the value of the first position to the right of the units digit is 5 of
Or
1 are
jo '1= 7
The value of the second position to the right of the units digit is — of =
or
eg : _1
a 100
The value of the third position to the right of the units digit is TT of me or
1 1 1 1
10 100 49% ~~ 1000
This pattern continues.
We can now see that if we were to write digits in positions to the right of
the units positions, those positions have values that are fractions. Not only
do the positions have fractional values, but the fractional values are all
powers of 10 (10,107, 10%, Bg a)
Decimal Fractions
Decimal Point, Decimal
If we are to write numbers with digits appearing to the right of the units
digit, we must have a way of denoting where the whole number part ends
and the fractional part begins. Mathematicians denote the separation point
of the units digit and the tenths digit by writing a decimal point. The word
decimal comes from the Latin prefix "deci" which means ten, and we use it
because we use a base ten number system. Numbers written in this form are
called decimal fractions, or more simply, decimals.
, :
2 & 3
3 : 3 s 8 3 3
2 Bg @ 3 2 £ & BE ¢
3 | a 5 an n pre 3 ~ 3 |
28 8 = = 8 & 8 = a § 58 5;
» Ee q a al i cal a He ofes
tk Eo
decimal point
Notice that decimal numbers have the suffix "th."
Decimal Fraction
A decimal fraction is a fraction in which the denominator is a power of 10.
The following numbers are examples of decimals.
1. 42.6
The 6 is in the tenths position.
42.6 = 42
2. 9.8014
The 8 is in the tenths position.
The 0 is in the hundredths position.
The 1 is in the thousandths position.
The 4 is in the ten thousandths position.
8014
3.093
The 9 is in the tenths position.
The 3 is in the hundredths position.
93
0.93 =
Note:Quite often a zero is inserted in front of a decimal point (in the
units position) of a decimal fraction that has a value less than one.
This zero helps keep us from overlooking the decimal point.
4.0.7
The 7 is in the tenths position.
Note:We can insert zeros to the right of the right-most digit in a
decimal fraction without changing the value of the number.
(chee = Sager ees
2 =0.7=07=2=5
Reading Decimal Fractions
Reading a Decimal Fraction
To read a decimal fraction,
1. Read the whole number part as usual. (If the whole number is less than
1, omit steps 1 and 2.)
2. Read the decimal point as the word "and."
3. Read the number to the right of the decimal point as if it were a whole
number.
4. Say the name of the position of the last digit.
Sample Set A
Read the following numbers.
Example:
6.8
6. , 8
|___]«—— tenths position
six and eight tenths
Note:Some people read this as "six point eight." This phrasing gets the
message across, but technically, "six and eight tenths" is the correct
phrasing.
Example:
14.116
14.11, 6
: L__}«—— thousandths position
fourteen and one hundred sixteen thousandths
Example:
0.0019
0.001 9
s L__|«-— ten thousandths position
nineteen ten thousandths
Example:
81
Eighty-one
In this problem, the indication is that any whole number is a decimal
fraction. Whole numbers are often called decimal numbers.
81 = 81.0
Practice Set A
Read the following decimal fractions.
Exercise:
Problem: 12.9
Solution:
twelve and nine tenths
Exercise:
Problem: 4.86
Solution:
four and eighty-six hundredths
Exercise:
Problem: 7.00002
Solution:
seven and two hundred thousandths
Exercise:
Problem: 0.030405
Solution:
thirty thousand four hundred five millionths
Writing Decimal Fractions
Writing a Decimal Fraction
To write a decimal fraction,
1. Write the whole number part.
2. Write a decimal point for the word "and."
3. Write the decimal part of the number so that the right-most digit
appears in the position indicated in the word name. If necessary, insert
zeros to the right of the decimal point in order that the right-most digit
appears in the correct position.
Sample Set B
Write each number.
Example:
Thirty-one and twelve hundredths.
The decimal position indicated is the hundredths position.
og heltZ
Example:
Two and three hundred-thousandths.
The decimal position indicated is the hundred thousandths. We'll need to
insert enough zeros to the immediate right of the decimal point in order to
locate the 3 in the correct position.
2.00003
Example:
Six thousand twenty-seven and one hundred four millionths.
The decimal position indicated is the millionths position. We'll need to
insert enough zeros to the immediate right of the decimal point in order to
locate the 4 in the correct position.
6,027.000104
Example:
Seventeen hundredths.
The decimal position indicated is the hundredths position.
eh 7,
Practice Set B
Write each decimal fraction.
Exercise:
Problem: Three hundred six and forty-nine hundredths.
Solution:
306.49
Exercise:
Problem: Nine and four thousandths.
Solution:
9.004
Exercise:
Problem: Sixty-one millionths.
Solution:
0.000061
Exercises
For the following three problems, give the decimal name of the position of
the given number in each decimal fraction.
Exercise:
1. 3.941
9 is in the position.
4 is in the position.
Problem: 1 is in the position.
Solution:
Tenths; hundredths, thousandths
Exercise:
17.1085
1 is in the position.
0 is in the position.
8 is in the position.
Problem: 5 is in the position.
Exercise:
652.3561927
9 is in the position.
Problem: 7 is in the position.
Solution:
Hundred thousandths; ten millionths
For the following 7 problems, read each decimal fraction by writing it.
Exercise:
Problem: 9.2
Exercise:
Problem: 8.1
Solution:
eight and one tenth
Exercise:
Problem: 10.15
Exercise:
Problem: 55.06
Solution:
fifty-five and six hundredths
Exercise:
Problem: 0.78
Exercise:
Problem: 1.904
Solution:
one and nine hundred four thousandths
Exercise:
Problem
: 10.00011
For the following 10 problems, write each decimal fraction.
Exercise:
Problem
: Three and twenty one-hundredths.
Solution:
3.20
Exercise:
Problem
Exercise:
Problem
: Fourteen and sixty seven-hundredths.
: One and eight tenths.
Solution:
1.8
Exercise:
Problem
Exercise:
Problem
: Sixty-one and five tenths.
: Five hundred eleven and four thousandths.
Solution:
511.004
Exercise:
Problem: Thirty-three and twelve ten-thousandths.
Exercise:
Problem: Nine hundred forty-seven thousandths.
Solution:
0.947
Exercise:
Problem: Two millionths.
Exercise:
Problem: Seventy-one hundred-thousandths.
Solution:
0.00071
Exercise:
Problem: One and ten ten-millionths.
Calculator Problems
For the following 10 problems, perform each division using a calculator.
Then write the resulting decimal using words.
Exercise:
Problem: 3—4
Solution:
seventy-five hundredths
Exercise:
Problem: 1—8
Exercise:
Problem: 4—10
Solution:
four tenths
Exercise:
Problem: 2—5
Exercise:
Problem: 4—25
Solution:
sixteen hundredths
Exercise:
Problem: 1—50
Exercise:
Problem: 3—16
Solution:
one thousand eight hundred seventy-five ten thousandths
Exercise:
Problem: 15—8
Exercise:
Problem: 11—20
Solution:
fifty-five hundredths
Exercise:
Problem: 9—40
Exercises for Review
Exercise:
Problem: ({link]) Round 2,614 to the nearest ten.
Solution:
2610
Exercise:
Problem: (({link]) Is 691,428,471 divisible by 3?
Exercise:
({link]) Determine the missing numerator.
e 3 — ss
Problem: <7 = =
Solution:
12
Exercise:
Problem: ((!ink]) Find of 4
Exercise:
Problem: ({link]) Find the value of
Solution:
25
gr 1
Converting a Decimal to a Fraction
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Fllis, Jr. This module discusses how to covert a decimal to a
fraction. By the end of the module students should be able to convert an
ordinary decimal and a complex decimal to a fraction.
Section Overview
e Converting an Ordinary Decimal to a Fraction
e Converting a Complex Decimal to a Fraction
Converting an Ordinary Decimal to a Fraction
We can convert a decimal fraction to a fraction, essentially, by saying it in
words, then writing what we say. We may have to reduce that fraction.
Sample Set A
Convert each decimal fraction to a proper fraction or a mixed number.
Example:
0.6
Ni tenths position
Reading: six tenths > —
3
Reduce: =
Example:
0.903
thousandths position
903
Reading: nine hundred three thousands > +557.
Example:
3.70370
27) 100.00000
81
190
189
100
81
190
189
61
Reading: eighteen and sixty-one hundredths > 18—55.
Example:
508.0005
ten thousandths position
senoeee
Reading: five hundred eight and five ten thousandths > 50875997 -
Reduce: 5085555.
Practice Set A
Convert the following decimals to fractions or mixed numbers. Be sure to
reduce.
Exercise:
Problem: 16.84
Solution:
165
Exercise:
Problem: 0.513
Solution:
513
1,000
Exercise:
Problem: 6,646.0107
Solution:
107
6,646 10,000
Exercise:
Problem: 1.1
Solution:
Converting A Complex Decimal to a Fraction
Complex Decimals
Numbers such as 0.11 + are called complex decimals. We can also convert
complex decimals to fractions.
Sample Set B
Convert the following complex decimals to fractions.
Example:
0.112
The — appears to occur in the thousands position, but it is referring to ~ of
a hundredth. So, we read 0.114 as "eleven and two-thirds hundredths."
2 11-3+2
0.122% = =
es 100 100
e5)
as 3
> 2
1
= 85) 00)
ee: oe |
of
ee ie
es 3
20
aie
60
Example:
4.006
Note that 4.0064 = 4+ .0064
a “a
4+ .0067 = 4 =P naa
25
Practice Set B
Convert each complex decimal to a fraction or mixed number. Be sure to
reduce.
Exercise:
Problem: 0.84
Solution:
oo|~I
Exercise:
Problem: 0.122
Solution:
31
250
Exercise:
Problem: 6.0052
Solution:
7
6 1,200
Exercise:
; 3
Problem: 18.1 i7
Solution:
Exercises
For the following 20 problems, convert each decimal fraction to a proper
fraction or a mixed number. Be sure to reduce.
Exercise:
Problem: 0.7
Solution:
ag
10
Exercise:
Problem: 0.1
Exercise:
Problem: 0.53
Solution:
53
100
Exercise:
Problem: 0.71
Exercise:
Problem: 0.219
Solution:
219
1,000
Exercise:
Problem: 0.811
Exercise:
Problem: 4.8
Solution:
4
45
Exercise:
Problem: 2.6
Exercise:
Problem: 16.12
Solution:
3
16 35
Exercise:
Problem
Exercise:
Problem
> 25.88
> 6.0005
Solution:
1
6 2,000
Exercise:
Problem
Exercise:
Problem
1200
> 16.125
Solution:
1
164
Exercise:
Problem
Exercise:
Problem:
20375
3.04
Solution:
1
335
Exercise:
Problem
2211875
Exercise:
Problem: 8.225
Solution:
9
8 z0
Exercise:
Problem: 1.0055
Exercise:
Problem: 9.99995
Solution:
19,999
9 20,000
Exercise:
Problem: 22.110
For the following 10 problems, convert each complex decimal to a fraction.
Exercise:
Problem: 0.75
Solution:
3
4
Exercise:
Problem: 0.012
Exercise:
Problem
: 2.164
Solution:
13
2 80
Exercise:
Problem
Exercise:
Problem
‘ 2
: 5.182
1
: 141124
Solution:
337
14 3,000
Exercise:
Problem
Exercise:
Problem
: 80.00112
1.402,
Solution:
129
1 320
Exercise:
Problem:
Exercise:
0.8
Problem: 1.94
Solution:
1
295
Exercise:
Problem: 1.722
Exercises for Review
Exercise:
Problem:
({link]) Find the greatest common factor of 70, 182, and 154.
Solution:
14
Exercise:
Problem:
({link]) Find the greatest common multiple of 14, 26, and 60.
Exercise:
iit ; 3.15 . 5
Problem: ([link]) Find the value of = - 73 + 4.
Solution:
EE
10
Exercise:
Problem: ({link]) Find the value of 52 + 8-5.
Exercise:
Problem:
({link]) In the decimal number 26.10742, the digit 7 is in what
position?
Solution:
thousandths
Rounding Decimals
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to round decimals. By the
end of the module students should be able to round a decimal number to a
specified position.
Section Overview
e Rounding Decimal Numbers
Rounding Decimal Numbers
We first considered the concept of rounding numbers in [link] where our
concern with rounding was related to whole numbers only. With a few
minor changes, we can apply the same rules of rounding to decimals.
To round a decimal to a particular position:
1. Mark the position of the round-off digit (with an arrow or check).
2. Note whether the digit to the immediate right of the marked digit is
a. less than 5. If so, leave the round-off digit unchanged.
b. 5 or greater. If so, add 1 to the round-off digit.
3. If the round-off digit is
a. to the right of the decimal point, eliminate all the digits to its
right.
b. to the left of the decimal point, replace all the digits between it
and the decimal point with zeros and eliminate the decimal point
and all the decimal digits.
Sample Set A
Round each decimal to the specified position. (The numbers in parentheses
indicate which step is being used.)
Example:
Round 32.116 to the nearest hundredth.
32.116
hundredths position
e 2b The digit immediately to the right is 6, and 6 > 5, so we add 1 to
the round-off digit:
BS al
e 3a The round-off digit is to the right of the decimal point, so we
eliminate all digits to its right.
Beet
The number 32.116 rounded to the nearest hundredth is 32.12.
Example:
Round 633.14216 to the nearest hundred.
e 1
633.14216
hundreds position
e 2a The digit immediately to the right is 3, and 3 < 5 so we leave the
round-off digit unchanged.
e 3b The round-off digit is to the left of 0, so we replace all the digits
between it and the decimal point with zeros and eliminate the decimal
point and all the decimal digits.
600
The number 633.14216 rounded to the nearest hundred is 600.
Example:
1,729.63 rounded to the nearest ten is 1,730.
Example:
1.0144 rounded to the nearest tenth is 1.0.
Example:
60.98 rounded to the nearest one is 61.
Sometimes we hear a phrase such as "round to three decimal places." This
phrase means that the round-off digit is the third decimal digit (the digit in
the thousandths position).
Example:
67.129 rounded to the second decimal place is 67.13.
Example:
67.129558 rounded to 3 decimal places is 67.130.
Practice Set A
Round each decimal to the specified position.
Exercise:
Problem: 4.816 to the nearest hundredth.
Solution:
4.82
Exercise:
Problem
: 0.35928 to the nearest ten thousandths.
Solution:
0.3593
Exercise:
Problem
: 82.1 to the nearest one.
Solution:
82
Exercise:
Problem
: 753.98 to the nearest hundred.
Solution:
800
Exercise:
Problem
: Round 43.99446 to three decimal places.
Solution:
43.994
Exercise:
Problem: Round 105.019997 to four decimal places.
Solution:
105.0200
Exercise:
Problem: Round 99.9999 to two decimal places.
Solution:
100.00
Exercises
For the first 10 problems, complete the chart by rounding each decimal to
the indicated positions.
Exercise:
Problem: 20.01071
Tenth Hundredth Thousandth Ten Thousandth
Solution:
Tenth
20.0
Exercise:
Problem
Tenth
Exercise:
Problem
Tenth
Hundredth
20.01
Sowo20L2
Hundredth
Chae,
Mises RAG SIZ =.
Hundredth
Solution:
Thousandth
20.011
Thousandth
Thousandth
Ten Thousandth
20.0107
Ten Thousandth
Ten Thousandth
Tenth Hundredth Thousandth
931.2 po 1.22 531.219
Exercise:
Problem: 36.109053
Tenth Hundredth Thousandth
36.1
Exercise:
Problem: 1.999994
Tenth Hundredth Thousandth
Solution:
Ten Thousandth
931.2188
Ten Thousandth
Ten Thousandth
Tenth Hundredth
2.0 2.00
Exercise:
Problem: 7.4141998
Tenth Hundredth
Exercise:
Problem: 0.000007
Tenth Hundredth
Solution:
Thousandth
2.000
Thousandth
7.414
Thousandth
Ten Thousandth
2.0000
Ten Thousandth
Ten Thousandth
Tenth Hundredth
0.0 0.00
Exercise:
Problem: 0.00008
Tenth Hundredth
Exercise:
Problem: 9.19191919
Tenth Hundredth
Solution:
Thousandth
0.000
Thousandth
Thousandth
Ten Thousandth
0.0000
Ten Thousandth
0.0001
Ten Thousandth
Tenth Hundredth Thousandth Ten Thousandth
D2 9.19 3.192 9.1919
Exercise:
Problem: 0.0876543
Tenth Hundredth Thousandth Ten Thousandth
Calculator Problems
For the following 5 problems, round 18.4168095 to the indicated place.
Exercise:
Problem: 3 decimal places.
Solution:
18.417
Exercise:
Problem: 1 decimal place.
Exercise:
Problem: 5 decimal places.
Solution:
18.41681
Exercise:
Problem: 6 decimal places.
Exercise:
Problem: 2 decimal places.
Solution:
18.42
Calculator Problems
For the following problems, perform each division using a calculator.
Exercise:
Problem: 4 ~ 3 and round to 2 decimal places.
Exercise:
Problem: 1 ~ 8 and round to 1 decimal place.
Solution:
0.1
Exercise:
Problem: 1 ~ 27 and round to 6 decimal places.
Exercise:
Problem: 51 ~~ 61 and round to 5 decimal places.
Solution:
0.83607
Exercise:
Problem: 3 ~ 16 and round to 3 decimal places.
Exercise:
Problem: 16 ~ 3 and round to 3 decimal places.
Solution:
5.300
Exercise:
Problem: 26 ~ 7 and round to 5 decimal places.
Exercises for Review
Exercise:
Problem: ({link]) What is the value of 2 in the number 421,916,017?
Solution:
Ten million
Exercise:
Problem: ({link]) Perform the division: 378 + 29.
Exercise:
Problem: ((link]) Find the value of 4+.
Solution:
256
Exercise:
Problem: ({link]) Convert — to a mixed number.
Exercise:
Problem: ({link]) Convert 3.16 to a mixed number fraction.
Solution:
4
355
Addition and Subtraction of Decimals
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Fllis, Jr. This module discusses how to add and subtract
decimals. By the end of the module students should understand the method
used for adding and subtracting decimals, be able to add and subtract
decimals and be able to use the calculator to add and subtract decimals.
Section Overview
e The Logic Behind the Method
e The Method of Adding and Subtracting Decimals
e Calculators
The Logic Behind the Method
Consider the sum of 4.37 and 3.22. Changing each decimal to a fraction, we
have
437 + 3 Performing the addition, we get
_ 4 37 9): _.. At00+37 3-100-+22
4.37 + 3.22 = 4795 + 3469 = 100 + 100
437 322
00 + “too
437+322
100
759.
100
= BY)
= 100
= seven and fifty-nine hundredths
7.59
Thus, 4.37 + 3.22 = 7.59.
The Method of Adding and Subtracting Decimals
When writing the previous addition, we could have written the numbers in
columns.
4.37
+3.22
7.09
This agrees with our previous result. From this observation, we can suggest
a method for adding and subtracting decimal numbers.
Method of Adding and Subtracting Decimals
To add or subtract decimals:
1. Align the numbers vertically so that the decimal points line up under
each other and the corresponding decimal positions are in the same
column.
2. Add or subtract the numbers as if they were whole numbers.
3. Place a decimal point in the resulting sum or difference directly under
the other decimal points.
Sample Set A
Find the following sums and differences.
Example:
9.813 + 2.140
9.813 The decimal points are aligned in the same column.
+2.140
11.953
Example:
841.0056 + 47.016 + 19.058
841.0056
47.016
+19.058
To insure that the columns align properly, we can write a 0 in the position
at the end of the numbers 47.016 and 19.058 without changing their values.
ih Go |
841.0056
47.0160
+ 19.0580
907.0796
Example:
1.314 — 0.58
1.314
—0.58 Write a0 in the thousandths position.
Example:
16.01 — 7.053
16.01
—7.053 Write a 0 in the thousandths position.
15 910 10
16.919
— 7.053
8.957
Example:
Find the sum of 6.88106 and 3.5219 and round it to three decimal places.
6.88106
+3.5219 Write a0 in the ten thousandths position.
11
6.88106
+ 3.52190
10.40296
We need to round the sum to the thousandths position. Since the digit in the
position immediately to the right is 9, and 9>5, we get
10.403
Example:
Wendy has $643.12 in her checking account. She writes a check for
$16.92. How much is her new account balance?
To find the new account balance, we need to find the difference between
643.12 and 16.92. We will subtract 16.92 from 643.12.
31211
643.12
— 16.92
626.20
After writing a check for $16.92, Wendy now has a balance of $626.20 in
her checking account.
Practice Set A
Find the following sums and differences.
Exercise:
Problem: 3.187 + 2.992
Solution:
6.179
Exercise:
Problem: 14.987 — 5.341
Solution:
9.646
Exercise:
Problem: 0.5261 + 1.0783
Solution:
1.6044
Exercise:
Problem: 1.06 — 1.0535
Solution:
0.0065
Exercise:
Problem: 16,521.07 + 9,256.15
Solution:
2531 7 see
Exercise:
Problem:
Find the sum of 11.6128 and 14.07353, and round it to two decimal
places.
Solution:
25.69
Calculators
The calculator can be useful for finding sums and differences of decimal
numbers. However, calculators with an eight-digit display cannot be used
when working with decimal numbers that contain more than eight digits, or
when the sum results in more than eight digits. In practice, an eight-place
decimal will seldom be encountered. There are some inexpensive
calculators that can handle 13 decimal places.
Sample Set B
Use a calculator to find each sum or difference.
Example:
42.0638 + 126.551
Display Reads
Type 42.0638 42.0638
Press zt 42.0638
Type 126.551 126.551
Press = 168.6148
The sum is 168.6148.
Example:
Find the difference between 305.0627 and 14.29667.
Display Reads
Type 305.0627 305.0627
Press — 305.0627
Type 14.29667 14.29667
Press = 290.76603
The difference is 290.76603
Example:
91.07 + 3,891.001786
Since 3,891.001786 contains more than eight digits, we will be unable to
use an eight-digit display calculator to perform this addition. We can,
however, find the sum by hand.
51.070000
3891.001786
3942.071786
The sum is 3,942.071786.
Practice Set B
Use a calculator to perform each operation.
Exercise:
Problem: 4.286 + 8.97
Solution:
13.256
Exercise:
Problem: 452.0092 — 392.558
Solution:
59.4512
Exercise:
Problem: Find the sum of 0.095 and 0.001862
Solution:
0.096862
Exercise:
Problem: Find the difference between 0.5 and 0.025
Solution:
0.475
Exercise:
Problem: Find the sum of 2,776.00019 and 2,009.00012.
Solution:
Since each number contains more than eight digits, using some
calculators may not be helpful. Adding these by “hand technology,”
we get 4,785.00031
Exercises
For the following 15 problems, perform each addition or subtraction. Use a
calculator to check each result.
Exercise:
Problem: 1.84 + 7.11
Solution:
8.95
Exercise:
Problem: 15.015 — 6.527
Exercise:
Problem: 11.842 + 28.004
Solution:
39.846
Exercise:
Problem
Exercise:
Problem
: 3.16 — 2.52
: 3.55267 + 8.19664
Solution:
11.74931
Exercise:
Problem
Exercise:
Problem
: 0.9162 — 0.0872
> 65.512 — 8.3005
Solution:
oA2115
Exercise:
Problem
Exercise:
Problem
: 761.0808 — 53.198
> 4.305 + 2.119 — 3.817
Solution:
2.607
Exercise:
Problem: 19.1161 + 27.8014 + 39.3161
Exercise:
Problem: 0.41276 — 0.0018 — 0.00011
Solution:
0.41085
Exercise:
Problem: 2.181 + 6.05 + 1.167 + 8.101
Exercise:
Problem
: 1.0031+6.013106+0.00018+0.0092+2.11
Solution:
9.135586
Exercise:
Problem
Exercise:
Problem
> 27+ 42 + 9.16 — 0.1761 + 81.6
: 10.28 + 11.111 + 0.86+ 5.1
Solution:
27391
For the following 10 problems, solve as directed. A calculator may be
useful.
Exercise:
Problem: Add 6.1121 and 4.916 and round to 2 decimal places.
Exercise:
Problem: Add 21.66418 and 18.00184 and round to 4 decimal places.
Solution:
39.6660
Exercise:
Problem: Subtract 5.2121 from 9.6341 and round to 1 decimal place.
Exercise:
Problem: Subtract 0.918 from 12.006 and round to 2 decimal places.
Solution:
11.09
Exercise:
Problem:
Subtract 7.01884 from the sum of 13.11848 and 2.108 and round to 4
decimal places.
Exercise:
Problem:
A checking account has a balance of $42.51. A check is written for
$19.28. What is the new balance?
Solution:
23.23
Exercise:
Problem:
A checking account has a balance of $82.97. One check is written for
$6.49 and another for $39.95. What is the new balance?
Exercise:
Problem:
A person buys $4.29 worth of hamburger and pays for it with a $10
bill. How much change does this person get?
Solution:
$5.71
Exercise:
Problem:
A man buys $6.43 worth of stationary and pays for it with a $20 bill.
After receiving his change, he realizes he forgot to buy a pen. If the
total price of the pen is $2.12, and he buys it, how much of the $20 bill
is left?
Exercise:
Problem:
A woman starts recording a movie on her video cassette recorder with
the tape counter set at 21.93. The movie runs 847.44 tape counter
units. What is the final tape counter reading?
Solution:
869.37
Exercises for Review
Exercise:
Problem: ({link]) Find the difference between 11,206 and 10,884.
Exercise:
Problem: ({link]) Find the product, 820 - 10,000.
Solution:
8,200,000
Exercise:
Problem: ({[link]) Find the value of 121 — 25 + 82 + 16 = 2?.
Exercise:
Problem: ([link]) Find the value of 85 . 3 + 22.
Solution:
20 _ 5 2
9 3 Or 25
Exercise:
Problem: ({link]) Round 1.08196 to the nearest hundredth.
Multiplication of Decimals
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to multiply decimals. By the
end of the module students should understand the method used for
multiplying decimals, be able to multiply decimals, be able to simplify a
multiplication of a decimal by a power of 10 and understand how to use the
word "of" in multiplication.
Section Overview
¢ The Logic Behind the Method
The Method of Multiplying Decimals
Calculators
Multiplying Decimals By Powers of 10
e Multiplication in Terms of “Of”
The Logic Behind the Method
Consider the product of 3.2 and 1.46. Changing each decimal to a fraction,
we have
(3.2) (1.46) Baad cag
100
32, 146
10 100
32-146
10-100
4672
1000
672
45060
= four and six hundred seventy-two thousandths
4.672
Thus, (3.2) (1.46) = 4.672.
Notice that the factor
3.2 has 1 decimal place,
1.46 has 2 decimal pl
as 2 decimal places, i723
and the product
4.672 has 3 decimal places.
Using this observation, we can suggest that the sum of the number of
decimal places in the factors equals the number of decimal places in the
product.
1
1.46<— 2 decimal places
X 3.2<— +1 decimal place
292
438
4.672 <—— 3 decimal places
The Method of Multiplying Decimals
Method of Multiplying Decimals
To multiply decimals,
1. Multiply the numbers as if they were whole numbers.
2. Find the sum of the number of decimal places in the factors.
3. The number of decimal places in the product is the sum found in step
2:
Sample Set A
Find the following products.
Example:
6.5-4.3
6,5 <— 1 decimal plate _ } ;
4.3, <— 1 decimal plac a 1 + 1 =2 decimal places in the product.
195
260
27,95 <— 2 decimal places
Thus, 6.5- 4.3 = 27.95.
Example:
23.4- 1.96
23.4 <— 1 decimal place
1.96 <—— 2 decimal places
1404
2106
234
45.864 <—— 3 decimal places
Thus, 23.4 - 1.96 = 45.864.
} 1 + 2 =3 decimal places in the product.
Thus, 23.4 - 1.96 = 45.864.
Example:
Find the product of 0.251 and 0.00113 and round to three decimal places.
0,251 <—— 3 decimal places
0.00113 <—— 5 decimal places
753
251
251
0.00028363
} 3 + 5 = 8 decimal places in the product.
We need to add three zeros to get 8 decimal places.
Now, rounding to three decimal places, we get
0.251 - 0.00113 = 0.000
to three decimal places.
Practice Set A
Find the following products.
Exercise:
Problem: 5.3 - 8.6
Solution:
45.58
Exercise:
Problem: 2.12 - 4.9
Solution:
10.388
Exercise:
Problem: 1.054 - 0.16
Solution:
0.16864
Exercise:
Problem: 0.00031 - 0.002
Solution:
0.00000062
Exercise:
Problem:
Find the product of 2.33 and 4.01 and round to one decimal place.
Solution:
eae)
Exercise:
Problem: 10 - 5.394
Solution:
93.94
Exercise:
Problem: 100 - 5.394
Solution:
939.4
Exercise:
Problem: 1000 - 5.394
Solution:
5,394
Exercise:
Problem: 10,000 - 5.394
Solution:
59,340
Calculators
Calculators can be used to find products of decimal numbers. However, a
calculator that has only an eight-digit display may not be able to handle
numbers or products that result in more than eight digits. But there are
plenty of inexpensive ($50 - $75) calculators with more than eight-digit
displays.
Sample Set B
Find the following products, if possible, using a calculator.
Example:
2.58 - 8.61
Display Reads
Type 2.58 2.58
Press x 2.58
Type 8.61 8.61
Press = 22.2138
The product is 22.2138.
Example:
0.006 - 0.0042
Display Reads
Type .006 .006
Press x 006
Type 0042 0.0042
Press = 0.0000252
We know that there will be seven decimal places in the product (since
3 + 4 = 7). Since the display shows 7 decimal places, we can assume the
product is correct. Thus, the product is 0.0000252.
Example:
0.0026 - 0.11976
Since we expect 4 + 5 = 9 decimal places in the product, we know that an
eight-digit display calculator will not be able to provide us with the exact
value. To obtain the exact value, we must use "hand technology." Suppose,
however, that we agree to round off this product to three decimal places.
We then need only four decimal places on the display.
Display Reads
Type .0026 0026
Press x 0026
Type .11976 0.11976
Press = 0.0003114
Rounding 0.0003114 to three decimal places we get 0.000. Thus,
0.0026 - 0.11976 = 0.000 to three decimal places.
Practice Set B
Use a calculator to find each product. If the calculator will not provide the
exact product, round the result to four decimal places.
Exercise:
Problem: 5.126 - 4.08
Solution:
20.91408
Exercise:
Problem: 0.00165 - 0.04
Solution:
0.000066
Exercise:
Problem: 0.5598 - 0.4281
Solution:
0.2397
Exercise:
Problem: 0.000002 - 0.06
Solution:
0.0000
Multiplying Decimals by Powers of 10
There is an interesting feature of multiplying decimals by powers of 10.
Consider the following multiplications.
Number
f Number of
a Positions the
Multiplication ee Decimal Point
ani Has Been Moved
of 10 to the Right
10 - 8.315274 = 83.15274 1 1
100 - 8.315274 = 831.5274 wi 2
1,000 - 8.315274 = 8,315.274 3 3
10,000 - 8.315274 = 83,152.74 4 4
Multiplying a Decimal by a Power of 10
To multiply a decimal by a power of 10, move the decimal place to the right
of its current position as many places as there are zeros in the power of 10.
Add zeros if necessary.
Sample Set C
Find the following products.
Example:
100 - 34.876. Since there are 2 zeros in 100, Move the decimal point in
34.876 two places to the right.
100 - 34.876 = 3487.6
= 3,487.6
Example:
1,000 - 4.8058. Since there are 3 zeros in 1,000, move the decimal point in
4.8058 three places to the right.
1,000 - 4.8058 = 4805.8
sai
= 4,805.8
Example:
10,000 - 56.82. Since there are 4 zeros in 10,000, move the decimal point
in 56.82 four places to the right. We will have to add two zeros in order to
obtain the four places.
10,000 - 56.82 = 568200.
LA
= 568,200
Since there is no fractional part, we can drop the decimal point.
Example:
(1,000,000)(2.57) = 2570000.
“S>_?
= 2,570,000
Example:
(1,000)(0.0000029) = 0 000.0029
4
= 0.0029
Practice Set C
Find the following products.
Exercise:
Problem: 100 - 4.27
Solution:
427
Exercise:
Problem: 10,000 - 16.52187
Solution:
165,218.7
Exercise:
Problem: (10) (0.0188)
Solution:
0.188
Exercise:
Problem: (10,000,000,000)(52.7)
Solution:
527,000,000,000
Multiplication in Terms of “Of”
Recalling that the word "of" translates to the arithmetic operation of
multiplication, let's observe the following multiplications.
Sample Set D
Example:
Find 4.1 of 3.8.
Translating "of" to "x", we get
|
x 3.8
328
123
15.58
Thus, 4.1 of 3.8 is 15.58.
Example:
Find 0.95 of the sum of 2.6 and 0.8.
We first find the sum of 2.6 and 0.8.
2.6
+0.8
34
Now find 0.95 of 3.4
oc
x 0.95
170
306
3.230
Thus, 0.95 of (2.6 + 0.8) is 3.230.
Practice Set D
Exercise:
Problem: Find 2.8 of 6.4.
Solution:
17.92
Exercise:
Problem: Find 0.1 of 1.3.
Solution:
0.13
Exercise:
Problem: Find 1.01 of 3.6.
Solution:
3.636
Exercise:
Problem: Find 0.004 of 0.0009.
Solution:
0.0000036
Exercise:
Problem: Find 0.83 of 12.
Solution:
206
Exercise:
Problem: Find 1.1 of the sum of 8.6 and 4.2.
Solution:
14.08
Exercises
For the following 30 problems, find each product and check each result
with a calculator.
Exercise:
Problem: 3.4 - 9.2
Solution:
31.28
Exercise:
Problem:4.5 - 6.1
Exercise:
Problem:8.0 - 5.9
Solution:
47.20
Exercise:
Problem:6.1 - 7
Exercise:
Problem: (0.1) (1.52)
Solution:
0.152
Exercise:
Problem: (1.99)(0.05)
Exercise:
Problem: (12.52) (0.37)
Solution:
4.6324
Exercise:
Problem: (5.116) (1.21)
Exercise:
Problem: (31.82) (0.1)
Solution:
3.182
Exercise:
Problem: (16.527) (9.16)
Exercise:
Problem:0.0021 - 0.013
Solution:
0.0000273
Exercise:
Problem: 1.0037 - 1.00037
Exercise:
Problem: (1.6)(1.6)
Solution:
2.56
Exercise:
Problem: (4.2) (4.2)
Exercise:
Problem:0.9 - 0.9
Solution:
0.81
Exercise:
Problem: 1.11 - 1.11
Exercise:
Problem:6.815 - 4.3
Solution:
29.3045
Exercise:
Problem: 9.0168 - 1.2
Exercise:
Problem: (3.5162) (0.0000003)
Solution:
0.00000105486
Exercise:
Problem: (0.000001)(0.01)
Exercise:
Problem: (10) (4.96)
Solution:
49.6
Exercise:
Problem: (10) (36.17)
Exercise:
Problem: 10 - 421.8842
Solution:
4,218.842
Exercise:
Problem: 10 - 8.0107
Exercise:
Problem: 100 - 0.19621
Solution:
19.621
Exercise:
Problem: 100 - 0.779
Exercise:
Problem: 1000 - 3.596168
Solution:
3,596.168
Exercise:
Problem: 1000 - 42.7125571
Exercise:
Problem: 1000 - 25.01
Solution:
25,010
Exercise:
Problem: 100,000 - 9.923
Exercise:
Problem: (4.6) (6.17)
Actual product Tenths Hundreds
Solution:
Actual product Tenths Hundreds
28.382 28.4 28.38
Exercise:
Problem: (8.09)(7.1)
Actual product Tenths Hundreds
Thousandths
Thousandths
28.382
Thousandths
Exercise:
Problem: (11.1106) (12.08)
Actual product Tenths Hundreds
Solution:
Actual product Tenths Hundreds
134.216048 134.2 134.22
Exercise:
Problem: 0.0083 - 1.090901
Actual product Tenths Hundreds
Thousandths
Thousandths
134.216
Thousandths
Exercise:
Problem: 7 - 26.518
Actual product Tenths Hundreds Thousandths
Solution:
Actual product Tenths Hundreds Thousandths
185.626 185.6 185.63 185.626
For the following 15 problems, perform the indicated operations
Exercise:
Problem:Find 5.2 of 3.7.
Exercise:
Problem:Find 12.03 of 10.1
Solution:
121.503
Exercise:
Problem:Find 16 of 1.04
Exercise:
Problem:Find 12 of 0.1
Solution:
1.2
Exercise:
Problem:Find 0.09 of 0.003
Exercise:
Problem:Find 1.02 of 0.9801
Solution:
0.999702
Exercise:
Problem:Find 0.01 of the sum of 3.6 and 12.18
Exercise:
Problem:Find 0.2 of the sum of 0.194 and 1.07
Solution:
0.2528
Exercise:
Problem:Find the difference of 6.1 of 2.7 and 2.7 of 4.03
Exercise:
Problem:Find the difference of 0.071 of 42 and 0.003 of 9.2
Solution:
2.9544
Exercise:
Problem:
If a person earns $8.55 an hour, how much does he earn in twenty-five
hundredths of an hour?
Exercise:
Problem:A man buys 14 items at $1.16 each. What is the total cost?
Solution:
$16.24
Exercise:
Problem:
In the problem above, how much is the total cost if 0.065 sales tax is
added?
Exercise:
Problem:
A river rafting trip is supposed to last for 10 days and each day 6 miles
is to be rafted. On the third day a person falls out of the raft after only
2. of that day’s mileage. If this person gets discouraged and quits, what
fraction of the entire trip did he complete?
Solution:
0.24
Exercise:
Problem:
A woman starts the day with $42.28. She buys one item for $8.95 and
another for $6.68. She then buys another item for sixty two-hundredths
of the remaining amount. How much money does she have left?
Calculator Problems
For the following 10 problems, use a calculator to determine each product.
If the calculator will not provide the exact product, round the results to five
decimal places.
Exercise:
Problem: 0.019 - 0.321
Solution:
0.006099
Exercise:
Problem: 0.261 - 1.96
Exercise:
Problem: 4.826 - 4.827
Solution:
23.295102
Exercise:
Problem: (9.46)
Exercise:
Problem: (0.012)
Solution:
0.000144
Exercise:
Problem: 0.00037 - 0.0065
Exercise:
Problem: 0.002 - 0.0009
Solution:
0.0000018
Exercise:
Problem: 0.1286 - 0.7699
Exercise:
Problem: 0.01 - 0.00000471
Solution:
0.0000000471
Exercise:
Problem: 0.00198709 - 0.03
Exercises for Review
Exercise:
Problem: ({link]) Find the value, if it exists, of 0 + 15.
Solution:
0
Exercise:
Problem:
({link]) Find the greatest common factor of 210, 231, and 357.
Exercise:
Problem: ({link]) Reduce sr to lowest terms.
Solution:
10
77
Exercise:
Problem:
({link]) Write "fourteen and one hundred twenty-one ten-thousandths,
using digits."
Exercise:
Problem:
({link]) Subtract 6.882 from 8.661 and round the result to two decimal
places.
Solution:
1.78
Division of Decimals
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to divide decimals. By the
end of the module students should understand the method used for dividing
decimals, be able to divide a decimal number by a nonzero whole number
and by another, nonzero, decimal number and be able to simplify a division
of a decimal by a power of 10.
Section Overview
The Logic Behind the Method
A Method of Dividing a Decimal By a Nonzero Whole Number
e A Method of Dividing a Decimal by a Nonzero Decimal
e Dividing Decimals by Powers of 10
The Logic Behind the Method
As we have done with addition, subtraction, and multiplication of decimals,
we will study a method of division of decimals by converting them to
fractions, then we will make a general rule.
We will proceed by using this example: Divide 196.8 by 6.
We have, up to this point, divided 196.8 by 6 and have gotten a quotient of
32 with a remainder of 4. If we follow our intuition and bring down the .8,
we have the division 4.8 ~ 6.
ABB as A SG
10
48. 6
_ 10 ° 1
8
2 #4
— 10 4
1
i, BE
=> 10
Thus, 4.8 + 6 = .8.
Now, our intuition and experience with division direct us to place the .8
immediately to the right of 32.
f Notice that the decimal points appear in the same column.
32.8
From these observations, we suggest the following method of division.
A Method of Dividing a Decimal by a Nonzero Whole Number
Method of Dividing a Decimal by a Nonzero Whole Number
To divide a decimal by a nonzero whole number:
1. Write a decimal point above the division line and directly over the
decimal point of the dividend.
2. Proceed to divide as if both numbers were whole numbers.
3. If, in the quotient, the first nonzero digit occurs to the right of the
decimal point, but not in the tenths position, place a zero in each
position between the decimal point and the first nonzero digit of the
quotient.
Sample Set A
Find the decimal representations of the following quotients.
Example:
MAT = 7
16.3
7)114.1
Thus, 114.1 + 7 = 16.3.
Check: If 114.1 + 7 = 16.3, then 7 - 16.3 should equal 114.1.
42
16.3
fi
114.1 True.
Example:
0.02068 + 4
Place zeros in the tenths and hundredths positions. (See Step 3.)
Thus, 0.02068 + 4 = 0.00517.
Practice Set A
Find the following quotients.
Exercise:
Problem: 184.5 = 3
Solution:
61.5
Exercise:
Problem: 16.956 = 9
Solution:
1.884
Exercise:
Problem: 0.2964 = 4
Solution:
0.0741
Exercise:
Problem: 0.000496 + 8
Solution:
0.000062
A Method of Dividing a Decimal By a Nonzero Decimal
Now that we can divide decimals by nonzero whole numbers, we are in a
position to divide decimals by a nonzero decimal. We will do so by
converting a division by a decimal into a division by a whole number, a
process with which we are already familiar. We'll illustrate the method
using this example: Divide 4.32 by 1.8.
Let's look at this problem as 4-32 = 1 -.
5) es 100
4500 “139
The divisor is 8. We can convert 8 into a whole number if we multiply it
by 10.
18 _ 1 OW
18.49 = 48. 2 18
1
But, we know from our experience with fractions, that if we multiply the
denominator of a fraction by a nonzero whole number, we must multiply the
numerator by that same nonzero whole number. Thus, when converting =>
to a whole number by multiplying it by 10, we must also multiply the
432
numerator 354 by 10.
432 49 — 432 We 4321 432
100 ~ yer 1 101 “10
10
2b feed
= 4375
=, AS?
We have converted the division 4.32 ~ 1.8 into the division 43.2 + 18, that
is,
1.8)4.32 — 18)43.2
Notice what has occurred.
1.8) 4.32 —> 18) 43.2
If we "move" the decimal point of the divisor one digit to the right, we must
also "move" the decimal point of the dividend one place to the right. The
word "move" actually indicates the process of multiplication by a power of
10.
Method of Dividing a Decimal by a Decimal Number
To divide a decimal by a nonzero decimal,
1. Convert the divisor to a whole number by moving the decimal point to
the position immediately to the right of the divisor's last digit.
2. Move the decimal point of the dividend to the right the same number
of digits it was moved in the divisor.
3. Set the decimal point in the quotient by placing a decimal point
directly above the newly located decimal point in the dividend.
4. Divide as usual.
Sample Set B
Find the following quotients.
Example:
32-06-71
7.1)32.66
e The divisor has one decimal place.
e¢ Move the decimal point of both the divisor and the dividend 1 place to
the right.
e Set the decimal point.
e Divide as usual.
Thus, 32.66 + 7.1 = 4.6.
Check: 32.66 + 7.1 = 4.6 if 4.6 x 7.1 = 32.66
4.6
pal fe |
46
o22
32.66 True.
Example:
1.0773 + 0.513
e The divisor has 3 decimal places.
¢ Move the decimal point of both the divisor and the dividend 3 places
to the right.
e Set the decimal place and divide.
US Os = Oral: 2d
Checking by multiplying 2.1 and 0.513 will convince us that we have
obtained the correct result. (Try it.)
Example:
12 + 0.00032
0.00032 )12.00000
e The divisor has 5 decimal places.
e Move the decimal point of both the divisor and the dividend 5 places
to the right. We will need to add 5 zeros to 12.
e Set the decimal place and divide.
0.00032 ) 12.00000
This is now the same as the division of whole numbers.
37500.
32) 1200000.
000
Checking assures us that 12 + 0.00032 = 37,500.
Practice Set B
Find the decimal representation of each quotient.
Exercise:
Problem: 9.176 ~ 3.1
Solution:
2:96
Exercise:
Problem
: 5.0838 + 1.11
Solution:
4.58
Exercise:
Problem
: 16 + 0.0004
Solution:
40,000
Exercise:
Problem
: 8,162.41 + 10
Solution:
816.241
Exercise:
Problem
: 8,162.41 + 100
Solution:
81.6241
Exercise:
Problem: 8,162.41 =~ 1,000
Solution:
8.16241
Exercise:
Problem: 8,162.41 + 10,000
Solution:
0.816241
Calculators
Calculators can be useful for finding quotients of decimal numbers. As we
have seen with the other calculator operations, we can sometimes expect
only approximate results. We are alerted to approximate results when the
calculator display is filled with digits. We know it is possible that the
operation may produce more digits than the calculator has the ability to
show. For example, the multiplication
0.12345 x 0.4567
5 decimal 4 decimal
places places
produces 5 + 4 = 9 decimal places. An eight-digit display calculator only
has the ability to show eight digits, and an approximation results. The way
to recognize a possible approximation is illustrated in problem 3 of the next
sample set.
Sample Set C
Find each quotient using a calculator. If the result is an approximation,
round to five decimal places.
Example:
12.596 + 4.7
Display Reads
Type 12.596 12.596
Press = 12.596
Type 4.7 4.7
Press = 2.68
Since the display is not filled, we expect this to be an accurate result.
Example:
0.5696376 + 0.00123
Display Reads
Type .0696376 0.5696376
Press Za 0.5696376
Type .00123 0.00123
Press = 463.12
Since the display is not filled, we expect this result to be accurate.
Example:
0.8215199 + 4.113
Display Reads
Type 8215199 0.8215199
Press z 0.8215199
Type 4.113 4.113
Press = 0.1997373
There are EIGHT DIGITS — DISPLAY FILLED! BE AWARE OF
POSSIBLE APPROXIMATIONS.
We can check for a possible approximation in the following way. Since the
3
division 4)12 can be checked by multiplying 4 and 3, we can check our
division by performing the multiplication
4.113 x 0.1997373
3 decimal 7 decimal
places places
This multiplication produces 3 + 7 = 10 decimal digits. But our suspected
quotient contains only 8 decimal digits. We conclude that the answer is an
approximation. Then, rounding to five decimal places, we get 0.19974.
Practice Set C
Find each quotient using a calculator. If the result is an approximation,
round to four decimal places.
Exercise:
Problem: 42.49778 ~ 14.261
Solution:
2.98
Exercise:
Problem: 0.001455 = 0.291
Solution:
0.005
Exercise:
Problem: 7.459085 = 2.1192
Solution:
3.5197645 is an approximate result. Rounding to four decimal places,
we get 3.5198
Dividing Decimals By Powers of 10
In problems 4 and 5 of [link], we found the decimal representations of
8,162.41 + 10 and 8,162.41 + 100. Let's look at each of these again and
then, from these observations, make a general statement regarding division
of a decimal number by a power of 10.
816.241
10)8162.410
Thus, 8,162.41 + 10 = 816.241.
Notice that the divisor 10 is composed of one 0 and that the quotient
816.241 can be obtained from the dividend 8,162.41 by moving the decimal
point one place to the left.
81.6241
100 )8162.4100
Thus, 8,162.41 + 100 = 81.6241.
Notice that the divisor 100 is composed of two 0's and that the quotient
81.6241 can be obtained from the dividend by moving the decimal point
two places to the left.
Using these observations, we can suggest the following method for dividing
decimal numbers by powers of 10.
Dividing a Decimal Fraction by a Power of 10
To divide a decimal fraction by a power of 10, move the decimal point of
the decimal fraction to the left as many places as there are zeros in the
power of 10. Add zeros if necessary.
Sample Set D
Find each quotient.
Example:
9,248.6 + 100
Since there are 2 zeros in this power of 10, we move the decimal point 2
places to the left.
92 48.6 + 100 = 92.486
KY
Example:
3.28 + 10,000
Since there are 4 zeros in this power of 10, we move the decimal point 4
places to the left. To do so, we need to add three zeros.
0003.28 + 10,000 = 0.000328
ad
Practice Set D
Find the decimal representation of each quotient.
Exercise:
Problem: 182.5 — 10
Solution:
18:25
Exercise:
Problem: 182.5 — 100
Solution:
1825
Exercise:
Problem: 182.5 ~ 1,000
Solution:
0.1825
Exercise:
Problem: 182.5 + 10,000
Solution:
0.01825
Exercise:
Problem: 646.18 — 100
Solution:
6.4618
Exercise:
Problem
: 21.926 + 1,000
Solution:
0.021926
Exercises
For the following 30 problems, find the decimal representation of each
quotient. Use a calculator to check each result.
Exercise:
Problem:
4.8+3
Solution:
1.6
Exercise:
Problem
Exercise:
Problem
: 16.8+8
2 18.5+5
Solution:
3.7
Exercise:
Problem
Exercise:
Problem
212.333
> 54.36+9
Solution:
6.04
Exercise:
Problem
Exercise:
Problem
> 73.56+12
: 159.46+17
Solution:
9.38
Exercise:
Problem
Exercise:
Problem
: 12.16+64
> 37.26+81
Solution:
0.46
Exercise:
Problem
Exercise:
Problem
> 439.35+435
> 36.98+4.3
Solution:
8.6
Exercise:
Problem
Exercise:
Problem
: 46.41+9.1
29.0155
Solution:
2.4
Exercise:
Problem
Exercise:
Problem
: 0.68+1.7
: 50.301+8.1
Solution:
6.21
Exercise:
Problem
Exercise:
Problem
> 2.832+0.4
> 4.7524+2.18
Solution:
2.18
Exercise:
Problem
Exercise:
Problem
: 16.2409+4.03
: 1.002001-+1.001
Solution:
1.001
Exercise:
Problem: 25.050025+5.005
Exercise:
Problem: 12.4—3.1
Solution:
4
Exercise:
Problem: 0.48—0.08
Exercise:
Problem: 30.24—2.16
Solution:
14
Exercise:
Problem: 48.87—0.87
Exercise:
Problem: 12.321—0.111
Solution:
iit
Exercise:
Problem: 64,351.006~+10
Exercise:
Problem: 64,351.006~+100
Solution:
643.51006
Exercise:
Problem: 64,351.0061,000
Exercise:
Problem: 64,351.006~1,000,000
Solution:
0.064351006
Exercise:
Problem: 0.43—100
For the following 5 problems, find each quotient. Round to the specified
position. A calculator may be used.
Exercise:
Problem: 11.2944—6.24
Actual Quotient Tenths Hundredths Thousandths
Solution:
Actual Quotient Tenths Hundredths
1.81 1.8 1.81
Exercise:
Problem: 45.32931—9.01
Actual Quotient Tenths Hundredths
Exercise:
Problem: 3.18186—0.66
Actual Quotient Tenths Hundredths
Thousandths
1.810
Thousandths
Thousandths
Solution:
Actual Quotient Tenths Hundredths
4.821 4.8 4.82
Exercise:
Problem: 4.3636—4
Actual Quotient Tenths Hundredths
Exercise:
Problem: 0.00006318-+-0.018
Actual Quotient Tenths Hundredths
Thousandths
4.821
Thousandths
Thousandths
Solution:
Actual Quotient Tenths Hundredths Thousandths
0.00351 0.0 0.00 0.004
For the following 9 problems, find each solution.
Exercise:
Problem: Divide the product of 7.4 and 4.1 by 2.6.
Exercise:
Problem:
Divide the product of 11.01 and 0.003 by 2.56 and round to two
decimal places.
Solution:
0.01
Exercise:
Problem:
Divide the difference of the products of 2.1 and 9.3, and 4.6 and 0.8 by
0.07 and round to one decimal place.
Exercise:
Problem:
A ring costing $567.08 is to be paid off in equal monthly payments of
$46.84. In how many months will the ring be paid off?
Solution:
12.11 months
Exercise:
Problem: Six cans of cola cost $2.58. What is the price of one can?
Exercise:
Problem:
A family traveled 538.56 miles in their car in one day on their
vacation. If their car used 19.8 gallons of gas, how many miles per
gallon did it get?
Solution:
27.2 miles per gallon
Exercise:
Problem:
Three college students decide to rent an apartment together. The rent is
$812.50 per month. How much must each person contribute toward the
rent?
Exercise:
Problem:
A woman notices that on slow speed her video cassette recorder runs
through 296.80 tape units in 10 minutes and at fast speed through
1098.16 tape units. How many times faster is fast speed than slow
speed?
Solution:
3.7
Exercise:
Problem:
A class of 34 first semester business law students pay a total of
$1,354.90, disregarding sales tax, for their law textbooks. What is the
cost of each book?
Calculator Problems
For the following problems, use calculator to find the quotients. If the result
is approximate (see Sample Set C [link]) round the result to three decimal
places.
Exercise:
Problem: 3.8994—2.01
Solution:
1.94
Exercise:
Problem: 0.067444—0.052
Exercise:
Problem: 14,115.628+484.74
Solution:
29,120
Exercise:
Problem: 219,709.36+9941.6
Exercise:
Problem: 0.0852092+0.49271
Solution:
O73
Exercise:
Problem: 2.4858225-+1.11611
Exercise:
Problem: 0.123432+0.1111
Solution:
1.111
Exercise:
Problem: 2.102838+1.0305
Exercises for Review
Exercise:
Problem: ({link]) Convert 4t to an improper fraction.
Solution:
39
8
Exercise:
Problem: ({link]) 2 of what number is £?
7 5
Exercise:
Problem: ({link]) Find the sum. 4 + 0 + 2.
Solution:
47
AT pve
39 OF 135
Exercise:
Problem: ({link]) Round 0.01628 to the nearest ten-thousandths.
Exercise:
Problem: ({link]) Find the product (2.06)(1.39)
Solution:
2.8634
Nonterminating Divisions
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses nonterminating divisions. By the
end of the module students should understand the meaning of a
nonterminating division and be able to recognize a nonterminating number
by its notation.
Section Overview
e Nonterminating Divisions
e Denoting Nonterminating Quotients
Nonterminating Divisions
Let's consider two divisions:
1983.0
2.4+3
Terminating Divisions
Previously, we have considered divisions like example 1, which is an
example of a terminating division. A terminating division is a division in
which the quotient terminates after several divisions (the remainder is zero).
Exact Divisions
The quotient in this problem terminates in the tenths position. Terminating
divisions are also called exact divisions.
Nonterminating Division
The division in example 2 is an example of a nonterminating division. A
non-terminating division is a division that, regardless of how far we carry
it out, always has a remainder.
1.333
3) 4.00000
Repeating Decimal
We can see that the pattern in the brace is repeated endlessly. Such a
decimal quotient is called a repeating decimal.
Denoting Nonterminating Quotients
We use three dots at the end of a number to indicate that a pattern repeats
itself endlessly.
4+3=1.333...
Another way, aside from using three dots, of denoting an endlessly
repeating pattern is to write a bar(_) above the repeating sequence of
digits.
4=-3=>13
The bar indicates the repeated pattern of 3.
Repeating patterns in a division can be discovered in two ways:
1. As the division process progresses, should the remainder ever be the
same as the dividend, it can be concluded that the division is
nonterminating and that the pattern in the quotient repeats. This fact is
illustrated in [link] of [link].
2. As the division process progresses, should the "product, difference"
pattern ever repeat two consecutive times, it can be concluded that the
division is nonterminating and that the pattern in the quotient repeats.
This fact is illustrated in [link] and 4 of [link].
Sample Set A
Carry out each division until the repeating pattern can be determined.
Example:
100 = 2%
3.70370
27) 100.00000
189
When the remainder is identical to the dividend, the division is
nonterminating. This implies that the pattern in the quotient repeats.
100 + 27 = 3.70370370... The repeating block is 703.
100 + 27 = 3.703
Example:
te
We see that this “product, difference”pattern repeats. We can conclude that
the division is nonterminating and that the quotient repeats.
he Die line repeatiiig piockis 4.
0M
Example:
Divide 2 by 11 and round to 3 decimal places.
Since we wish to round the quotient to three decimal places, we'll carry out
the division so that the quotient has four decimal places.
.1818
The number .1818 rounded to three decimal places is .182. Thus, correct to
three decimal places,
2~+11= 0.182
Example:
Divide 1 by 6.
We see that this “product, difference” pattern repeats. We can conclude that
the division is nonterminating and that the quotient repeats at the 6.
10 — 0516
Practice Set A
Carry out the following divisions until the repeating pattern can be
determined.
Exercise:
Problem: 1 — 3
Solution:
0.3
Exercise:
Problem: 5 — 6
Solution:
0.83
Exercise:
Problem
:11+9
Solution:
1.2
Exercise:
Problem:
Le 9
Solution:
1.8
Exercise:
Problem
: Divide 7 by 6 and round to 2 decimal places.
Solution:
LA7
Exercise:
Problem
: Divide 400 by 11 and round to 4 decimal places.
Solution:
36.3636
Exercises
For the following 20 problems, carry out each division until the repeating
pattern is determined. If a repeating pattern is not apparent, round the
quotient to three decimal places.
Exercise:
Problem: 4 — 9
Solution:
0.4
Exercise:
Problem: 8 — 11
Exercise:
Problem: 4 — 25
Solution:
0.16
Exercise:
Problem: 5 — 6
Exercise:
Problem: 1 — 7
Solution:
0.142857
Exercise:
Problem: 3 — 1.1
Exercise:
Problem: 20 — 1.9
Solution:
10.526
Exercise:
Problem
Exercise:
Problem
SO sont
oe Bs ee
Solution:
0.112
Exercise:
Problem
Exercise:
Problem
28.08 4-301:
col 8.2
Solution:
6.21951
Exercise:
Problem
Exercise:
Problem
: 0.213 + 0.31
: 0.009 + 1.1
Solution:
0.0081
Exercise:
Problem
Exercise:
Problem
Solution:
0.835
Exercise:
Problem
Exercise:
Problem
Solution:
0.3
Exercise:
Problem
Exercise:
Problem
16.03 = 1.9
: 0.518 + 0.62
71.55 + 0.27
: 0.333 + 0.999
: 0.444 + 0.999
> 0.555 + 0.27
Solution:
2.05
Exercise:
Problem: 3.8 — 0.99
Calculator Problems
For the following 10 problems, use a calculator to perform each division.
Exercise:
Problem: 7 — 9
Solution:
0.7
Exercise:
Problem: 8 — 11
Exercise:
Problem: 14 — 27
Solution:
0.518
Exercise:
Problem: 1 — 44
Exercise:
Problem: 2 — 44
Solution:
0.045
Exercise:
Problem: 0.7 + 0.9 (Compare this with [link].)
Exercise:
Problem: 80 ~ 110 (Compare this with [link].)
Solution:
0.72
Exercise:
Problem: 0.0707 + 0.7070
Exercise:
Problem: 0.1414 = 0.2020
Solution:
0.7
Exercise:
Problem: 1 + 0.9999999
Exercise for Review
Exercise:
Problem:
({link]) In the number 411,105, how many ten thousands are there?
Solution:
1
Exercise:
Problem: ({link]) Find the quotient, if it exists. 17 + 0.
Exercise:
Problem: ([link]) Find the least common multiple of 45, 63, and 98.
Solution:
4410
Exercise:
Problem:
({link]) Subtract 8.01629 from 9.00187 and round the result to three
decimal places.
Exercise:
Problem: ({link]) Find the quotient. 104.06 + 12.1.
Solution:
8.6
Converting a Fraction to a Decimal
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to convert a fraction to a
decimal. By the end of the module students should be able to convert a
fraction to a decimal.
Now that we have studied and practiced dividing with decimals, we are also
able to convert a fraction to a decimal. To do so we need only recall that a
fraction bar can also be a division symbol. Thus, o not only means "3
objects out of 4," but can also mean "3 divided by 4."
Sample Set A
Convert the following fractions to decimals. If the division is
nonterminating, round to two decimal places.
Example:
3 Divide 3 by 4.
75
Example:
= Divide 1 by 5.
Example:
2.. Divide 5 by 6.
ating.
2 = 0.833 --- We are to round to two decimal places.
Thus, 2 = 0.83 to two decimal places.
|
Example:
be. Note that b= =5-+ 7
Convert - to a decimal.
e125
8)1.000
huss 5 — Oe Ot ee 12D,
Example:
0.164 . This is a complex decimal.
Note that the 6 is in the hundredths position.
The number 0.164 is read as "sixteen and one-fourth hundredths."
This recurring remainder indicates that the division is nontermin-
Now, convert 42 to a decimal.
80
.1625
Thus, 0.164 = 0.1625.
Practice Set A
Convert the following fractions and complex decimals to decimals (in
which no proper fractions appear). If the divison is nonterminating, round to
two decimal places.
Exercise:
Problem:
eS
Solution:
0.25
Exercise:
Problem: 35
Solution:
0.04
Exercise:
Problem:
a)
Solution:
0.17
Exercise:
. 15
Problem: ie
Solution:
0.9375
Exercise:
Problem: 0.95
Solution:
O95
Exercise:
Problem: 8.01262
Solution:
8.0126375
Exercises
For the following 30 problems, convert each fraction or complex decimal
number to a decimal (in which no proper fractions appear).
Exercise:
Problem:
tole
Solution:
0.5
Exercise:
ons
Problem:
Exercise:
Problem:
oo|~a
Solution:
0.875
Exercise:
Problem:
Colon
Exercise:
Problem:
ones
Solution:
0.6
Exercise:
onfro
Problem:
Exercise:
Problem:
Solution:
0.04
Exercise:
Problem:
Exercise:
Problem: =~
Solution:
0.05
Exercise:
Problem:
Exercise:
Problem:
Solution:
0.02
Exercise:
Problem:
Exercise:
Problem:
Solution:
0.3
Exercise:
Problem:
Exercise:
Problem:
Solution:
0.1875
Exercise:
Problem:
Exercise:
Problem:
Solution:
0.037
Exercise:
Problem:
Exercise:
oo)
DOr
Problem: ——
Solution:
0.538461
Exercise:
Problem:
Exercise:
Problem: 72
Solution:
£6
Exercise:
Problem: 8—-
Exercise:
Problem: 1—-
Solution:
Lis
Exercise:
Problem: 65 ay
Exercise:
Problem: 101 5>
Solution:
101.24
Exercise:
Problem: 0.1
dole
Exercise:
Problem: 0.24;
Solution:
0.24125
Exercise:
Problem: 5.66 +
Exercise:
Problem: 810.3106 2
Solution:
810.31063125
Exercise:
Problem: 4.15
For the following 18 problems, convert each fraction to a decimal. Round to
five decimal places.
Exercise:
Problem:
Solution:
0.11111
Exercise:
Problem:
Exercise:
Problem:
Solution:
0.35333
Exercise:
Problem:
Exercise:
Problem:
Solution:
0.55556
Exercise:
Problem:
Exercise:
cole
col
Coles
cO|
colon
co|m>
Problem:
Solution:
0.77778
Exercise:
Problem:
Exercise:
Problem:
Solution:
0.09091
Exercise:
Problem:
Exercise:
Problem:
Solution:
27273
Exercise:
Problem:
Exercise:
co|N
=
aril
—
FI
—_
le
—
=
_
len
Problem:
Solution:
0.45455
Exercise:
_
Ble
Problem:
Exercise:
Problem:
_
Bln
Solution:
0.63636
Exercise:
Problem:
_
[20
Exercise:
—
|
Problem:
Solution:
0.81818
Exercise:
Problem: —
Calculator Problems
For the following problems, use a calculator to convert each fraction to a
decimal. If no repeating pattern seems to exist, round to four decimal
places.
Exercise:
16
125
Problem:
Solution:
0.128
Exercise:
Problem: =
Exercise:
Problem:
Solution:
0.9746
Exercise:
Problem:
Exercise:
Problem:
Solution:
0.0002
Exercise:
81,426
106,001
Problem:
Exercise:
16,501
426
Problem:
Solution:
38.7347
Exercises for Review
Exercise:
Problem: ({link]) Round 2,105,106 to the nearest hundred thousand.
Exercise:
Problem: ({link}) — of what number is 3?
Solution:
15
16
Exercise:
Problem: ({link]) Arrange ae i and if in increasing order.
Exercise:
Problem: ({link]) Convert the complex decimal 3.62 to a fraction.
Solution:
33 Or 3.725
Exercise:
Problem: ({link]) Find the quotient. 30 + 1.1.
Combinations of Operations with Decimals and Fractions
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This
module discusses combinations of operations with decimals and fractions. By the end of the
module students should be able to combine operations with decimals.
Having considered operations with decimals and fractions, we now consider operations that
involve both decimals and fractions.
Sample Set A
Perform the following operations.
Example:
0.38 - +: Convert both numbers to decimals or both numbers to fractions. We’ll convert to
decimals.
.25
4)1.00
es
20
20
0
To convert + to a decimal, divide 1 by 4.
Now multiply 0.38 and .25.
1
4
38
X.25
190
76_
.0950
Thus, 0.38 - + = 0.095.
In the problems that follow, the conversions from fraction to decimal, or decimal to fraction, and
some of the additions, subtraction, multiplications, and divisions will be left to you.
Example:
1.85 + — - 4.1 Convert - to a decimal.
1.85 + 0.375 - 4.1 Multiply before adding.
1.85 + 1.5375 Now add.
3.3875
Example:
( _ 0.28) Convert 0.28 to a fraction.
eee
5 (4 _ 28) = 2(2- 2)
13 \5 100 1 ie 225
Example:
ue +2--0.1211 = M2 +4 -0.1211
=
oo) |SIS
or
|
Slee Go|r cole [onli
| Co
--
S|F
S)
—
bo
—
—
3 2 net)
= a5 oy UZ — OL eEL
— elbo2 — ad
= 0.03515
3515
= {00,000
703
20,000
Practice Set A
Perform the following operations.
Exercise:
Problem: 2 +1.6
Solution:
1
2.2 Or a=
Exercise:
Problem: 8.91 + + - 1.6
Solution:
223
Convert this to fraction form
Exercise:
Problem
Solution:
10
Exercise:
Problem:
Solution:
12 (6.12+ 4)
0.156
yal
15
— 0.05
35 OF 0.04
Exercises
Exercise:
Problem:
Solution:
if
Exercise:
Problem:
Exercise:
Problem:
Solution:
0.112
Exercise:
Problem
Exercise:
Problem:
3
3 +0.7
i
= +0.1
B)
5 = 0.518
0.418 — $o
0.22 -
ee
Solution:
0.055
Exercise:
Problem: = - 8.4
onfeo
Exercise:
Problem: = - 3.19
Solution:
0.1276
Exercise:
Problem: ;;-> 0.05
Exercise:
Problem: & + 0.25
Solution:
0.7
Exercise:
Problem: 1+ 0.9-0.12
Exercise:
Problem: 9.26 + + - 0.81
Solution:
9.4625
Exercise:
Problem: 0.588 + - - 0.24
Exercise:
it 3
Problem: on 3.62 - =
Solution:
1.4075
Exercise:
Problem:
Exercise:
Problem:
Solution:
0.1875
Exercise:
Problem:
Exercise:
Problem:
Solution:
0.75
Exercise:
Problem:
Exercise:
Problem:
Solution:
0.615
Exercise:
Problem:
Exercise:
7+0.15+ +
(Gh - 05)
0.2 + (gy + 1.1143)
3. (0.875 + t)
5.198 — 0.26 - (
0.54 + (0.3)
2 1
(1.4)" — 1.65
14
3h + 0.119
Problem: (2) — 0.000625 + (1.1)?
)
Solution:
135
Exercise:
Problem: (0.6)" . (4 7
Exercise:
2
Problem: (5) — 0.125
Solution:
0.125
Exercise:
. 0.75 5
Problem: re +2
Exercise:
et 876.4
Problem: ( a +)
Solution:
0.15
Exercise:
oy
Problem: 8 - (sé i 2)
Exercise:
0.32,
12
Problem: 5s
Solution:
2.6
Exercise:
oroptem: AVS) o®
roblem:
Exercises for Review
Exercise:
Problem: ((link]) Is 21,480 divisible by 3?
Solution:
yes
Exercise:
Problem: ({link]) Expand 14*. Do not find the actual value.
Exercise:
Problem: ((link]) Find the prime factorization of 15,400.
Solution:
OF 252 Te U1
Exercise:
Problem: ((link]) Convert 8.016 to a fraction.
Exercise:
Problem: ((link]) Find the quotient. 16 + 27.
Solution:
0.592
Summary of Key Concepts
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module reviews the key concepts from the chapter
"Decimals."
Summary of Key Concepts
Decimal Point ([link])
A decimal point is a point that separates the units digit from the tenths digit.
Decimal or Decimal Fraction ([link])
A decimal fraction is a fraction whose denominator is a power of ten.
Converting a Decimal to a Fraction ((link])
Decimals can be converted to fractions by saying the decimal number in
words, then writing what was said.
Rounding Decimals ((link])
Decimals are rounded in much the same way whole numbers are rounded.
Addition and Subtraction of Decimals ([link])
To add or subtract decimals,
1. Align the numbers vertically so that the decimal points line up under
each other and the corresponding decimal positions are in the same
column.
2. Add or subtract the numbers as if they were whole numbers.
3. Place a decimal point in the resulting sum directly under the other
decimal points.
Multiplication of Decimals ({link])
To multiply two decimals,
1. Multiply the numbers as if they were whole numbers.
2. Find the sum of the number of decimal places in the factors.
3. The number of decimal places in the product is the number found in
step 2.
Multiplying Decimals by Powers of 10 ({link])
To multiply a decimal by a power of 10, move the decimal point to the right
as many places as there are zeros in the power of ten. Add zeros if
necessary.
Division of a Decimal by a Decimal ({link])
To divide a decimal by a nonzero decimal,
1. Convert the divisor to a whole number by moving the decimal point
until it appears to the right of the divisor's last digit.
2. Move the decimal point of the dividend to the right the same number
of digits it was moved in the divisor.
3. Proceed to divide.
4. Locate the decimal in the answer by bringing it straight up from the
dividend.
Dividing Decimals by Powers of 10 ({link])
To divide a decimal by a power of 10, move the decimal point to the left as
many places as there are zeros in the power of ten. Add zeros if necessary.
Terminating Divisions ({link])
A terminating division is a division in which the quotient terminates after
several divisions. Terminating divisions are also called exact divisions.
Nonterminating Divisions (({link])
A nonterminating division is a division that, regardless of how far it is
carried out, always has a remainder. Nonterminating divisions are also
called nonexact divisions.
Converting Fractions to Decimals ((link])
A fraction can be converted to a decimal by dividing the numerator by the
denominator.
Exercise Supplement
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is an exercise supplement for the chapter
"Decimals" and contains many exercise problems. Odd problems are
accompanied by solutions.
Exercise Supplement
Reading and Writing Decimals ([{link])
Exercise:
Problem:
The decimal digit that appears two places to the right of the decimal
point is in the position.
Solution:
hundredths
Exercise:
Problem:
The decimal digit that appears four places to the right of the decimal
point is in the position.
For problems 3-8, read each decimal by writing it in words.
Exercise:
Problem: 7.2
Solution:
seven and two tenths
Exercise:
Problem: 8.105
Exercise:
Problem: 16.52
Solution:
sixteen and fifty-two hundredths
Exercise:
Problem: 5.9271
Exercise:
Problem: 0.005
Solution:
five thousandths
Exercise:
Problem: 4.01701
For problems 9-13, write each decimal using digits.
Exercise:
Problem: Nine and twelve-hundredths.
Solution:
oie
Exercise:
Problem: Two and one hundred seventy-seven thousandths.
Exercise:
Problem: Fifty-six and thirty-five ten-thousandths.
Solution:
56.0035
Exercise:
Problem: Four tenths.
Exercise:
Problem: Four thousand eighty-one millionths.
Solution:
0.004081
Converting a Decimal to a Fraction ((link])
For problem 14-20, convert each decimal to a proper fraction or a mixed
number.
Exercise:
Problem: 1.07
Exercise:
Problem: 58.63
Solution:
63
85 755
Exercise:
Problem: 0.05
Exercise:
Problem: 0.144
Solution:
ii,
75
Exercise:
Problem: 1.09=
Exercise:
. 1
Problem: 4.01 a7
Solution:
14
4 oe
Exercise:
Problem: 9.114
Rounding Decimals ((link])
For problems 21-25, round each decimal to the specified position.
Exercise:
Problem: 4.087 to the nearest hundredth.
Solution:
4.09
Exercise:
Problem: 4.087 to the nearest tenth.
Exercise:
Problem: 16.5218 to the nearest one.
Solution:
17
Exercise:
Problem: 817.42 to the nearest ten.
Exercise:
Problem: 0.9811602 to the nearest one.
Solution:
1
Addition, Subtraction, Multiplication and Division of Decimals, and
Nonterminating Divisions ([{link],[link],[link],[link])
For problem 26-45, perform each operation and simplify.
Exercise:
Problem: 7.10 + 2.98
Exercise:
Problem: 14.007 — 5.061
Solution:
8.946
Exercise:
Problem: 1.2 - 8.6
Exercise:
Problem: 41.8 - 0.19
Solution:
7.942
Exercise:
Problem: 57.51 ~ 2.7
Exercise:
Problem: 0.54003 ~ 18.001
Solution:
0.03
Exercise:
Problem: 32,051.3585 + 23,006.9999
Exercise:
Problem: 100 - 1,816.001
Solution:
181,600.1
Exercise:
Problem: 1,000 - 1,816.001
Exercise:
Problem: 10.000 - 0.14
Solution:
1.4
Exercise:
Problem: 0.135888 + 16.986
Exercise:
Problem: 150.79 = 100
Solution:
1.5079
Exercise:
Problem: 4.119 ~ 10,000
Exercise:
Problem: 42.7 ~ 18
Solution:
2.312
Exercise:
Problem
Exercise:
Problem
:6.9 + 12
: 0.014 + 47.6. Round to three decimal places.
Solution:
0.000
Exercise:
Problem
Exercise:
Problem
: 8.8 + 19. Round to one decimal place.
:1.1+9
Solution:
0.12
Exercise:
Problem
Exercise:
Problem
V1.9
> 30+ 11.1
Solution:
2.702
Converting a Fraction to a Decimal ((link])
For problems 46-55, convert each fraction to a decimal.
Exercise:
Problem:
00 |eo
Exercise:
43
100
Problem:
Solution:
0.43
Exercise:
Problem: TGag.
Exercise:
Problem: 92
Solution:
9.571428
Exercise:
° i
Problem: 8 iG
Exercise:
Problem: 1.3
eo)
Solution:
1.3
Exercise:
Problem: 25.6 +
Exercise:
Problem: 125.125
Solution:
125.125125 (not repeating)
Exercise:
Problem: 9.114
Exercise:
Problem: 0.02
Solution:
0.083
Combinations of Operations with Decimals and Fractions ([link])
For problems 56-62, perform each operation.
Exercise:
Problem: 2 - 0.25
Exercise:
Problem:
Solution:
0.255
Exercise:
Problem:
Exercise:
Problem:
Solution:
9.425
Exercise:
Problem:
Exercise:
Problem:
Solution:
0.09343
Exercise:
on|oo
to|~N
19.375 + (4.375 — 1-4)
15,
602
- 1.36
(4 +1.75)
- (4 +0.30)
2.6+ 34
Problem: 443 + (54 +33)
Proficiency Exam
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is a proficiency exam to the chapter
"Decimals." Each problem is accompanied with a reference link pointing
back to the module that discusses the type of problem demonstrated in the
question. The problems in this exam are accompanied by solutions.
Proficiency Exam
Exercise:
Problem:
({link]) The decimal digit that appears three places to the right of the
decimal point is in the position.
Solution:
thousandth
Exercise:
Problem: ((link]) Write, using words, 15.036.
Solution:
fifteen and thirty-six thousandths
Exercise:
Problem:
({link]) Write eighty-one and twelve hundredths using digits. 81.12
Solution:
81.12
Exercise:
Problem:
({link]) Write three thousand seventeen millionths using digits.
Solution:
0.003017
Exercise:
Problem: ({link]) Convert 0.78 to a fraction. Reduce.
Solution:
39
50
Exercise:
Problem: ({link]) Convert 0.875 to a fraction. Reduce.
Solution:
ir
8
Exercise:
Problem: ({link]) Round 4.8063 to the nearest tenth.
Solution:
4.8
Exercise:
Problem: ([link]) Round 187.51 to the nearest hundred.
Solution:
200
Exercise:
Problem: ({link]) Round 0.0652 to the nearest hundredth.
Solution:
0.07
For problems 10-20, perform each operation.
Exercise:
Problem: ((link]) 15.026 + 5.971
Solution:
20.997
Exercise:
Problem: ({link]) 72.15 — 26.585
Solution:
45.565
Exercise:
Problem: ({link]) 16.2 - 4.8
Solution:
77.76
Exercise:
Problem: ({link]) 10,000 - 0.016
Solution:
16
Exercise:
Problem: ((link]) 44.64 + 18.6
Solution:
2.4
Exercise:
Problem: ({link]) 0.21387 + 0.19
Solution:
1.1256
Exercise:
Problem: ((link]) 0.27 — +
Solution:
0
Exercise:
Problem: ({link]) Convert 64 to a decimal.
Solution:
6.18
Exercise:
Problem: ({link]) Convert 0.5=2- to a decimal.
Solution:
0.055625
Exercise:
Problem: ({link]) 3; + 2.325
Solution:
5.45
Exercise:
Problem: ((link]) 2 x 0.5625
Solution:
2 or 0.2109375
Objectives
This module contains the learning objectives for the chapter "Ratios and
Rates" from Fundamentals of Mathematics by Denny Burzynski and Wade
Ellis, jr.
After completing this chapter, you should
Ratios and Rates ({link])
¢ be able to distinguish between denominate and pure numbers and
between ratios and rates
Proportions ({link])
¢ be able to describe proportions and find the missing factor in a
proportion
e be able to work with proportions involving rates
Applications of Proportions ({link])
e solve proportion problems using the five-step method
Percent ({link])
e understand the relationship between ratios and percents
e be able to make conversions between fractions, decimals, and percents
Fractions of One Percent ({link])
e understand the meaning of a fraction of one percent
e be able to make conversions involving fractions of one percent
Applications of Percents ({link])
e be able to distinguish between base, percent, and percentage
e be able to find the percentage, the percent, and the base
Ratios and Rates
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses ratios and rates. By the end of the
module students should be able to distinguish between denominate and pure
numbers and between ratios and rates.
Section Overview
e Denominate Numbers and Pure Numbers
e Ratios and Rates
Denominate Numbers and Pure Numbers
Denominate Numbers, Like and Unlike Denominate Numbers
It is often necessary or convenient to compare two quantities. Denominate
numbers are numbers together with some specified unit. If the units being
compared are alike, the denominate numbers are called like denominate
numbers. If units are not alike, the numbers are called unlike denominate
numbers. Examples of denominate numbers are shown in the diagram:
8 gallons 32 cents 54 miles
ns Os
The denominations
Pure Numbers
Numbers that exist purely as numbers and do not represent amounts of
quantities are called pure numbers. Examples of pure numbers are 8, 254,
0, 212, 2, and 0.07.
Numbers can be compared in two ways: subtraction and division.
Comparing Numbers by Subtraction and Division
Comparison of two numbers by subtraction indicates how much more
one number is than another.
Comparison by division indicates how many times larger or smaller one
number is than another.
Comparing Pure or Like Denominate Numbers by Subtraction
Numbers can be compared by subtraction if and only if they both are like
denominate numbers or both pure numbers.
Sample Set A
Example:
Compare 8 miles and 3 miles by subtraction.
8 mile — 3 miles = 5 miles
This means that 8 miles is 5 miles more than 3 miles.
Examples of use: I can now jog 8 miles whereas I used to jog only 3 miles.
So, I can now jog 5 miles more than I used to.
Example:
Compare 12 and 5 by subtraction.
12 — sh
This means that 12 is 7 more than 5.
Example:
Comparing 8 miles and 5 gallons by subtraction makes no sense.
8 miles — 5 gallons = ?
Example:
Compare 36 and 4 by division.
36 +4=9
This means that 36 is 9 times as large as 4. Recall that 36 + 4 = 9 can be
expressed as 20. =o)
Example:
Compare 8 miles and 2 miles by division.
8 miles __ 4
2 miles
This means that 8 miles is 4 times as large as 2 miles.
Example of use: I can jog 8 miles to your 2 miles. Or, for every 2 miles that
you jog, I jog 8. So, I jog 4 times as many miles as you jog.
Notice that when like quantities are being compared by division, we drop
the units. Another way of looking at this is that the units divide out
(cancel).
Example:
Compare 30 miles and 2 gallons by division.
30 miles 15 miles
2 gallons 1 gallon
Example of use: A particular car goes 30 miles on 2 gallons of gasoline.
This is the same as getting 15 miles to 1 gallon of gasoline.
Notice that when the quantities being compared by division are unlike
quantities, we do not drop the units.
Practice Set A
Make the following comparisons and interpret each one.
Exercise:
Problem: Compare 10 diskettes to 2 diskettes by
a. subtraction:
b. division:
Solution:
a. 8 diskettes; 10 diskettes is 8 diskettes more than 2 diskettes.
b. 5; 10 diskettes is 5 times as many diskettes as 2 diskettes.
Exercise:
Problem: Compare, if possible, 16 bananas and 2 bags by
a. subtraction:
b. division:
Solution:
a. Comparison by subtraction makes no sense.
bi 36 bananas = ae 8 bananas per bag.
Ratios and Rates
Ratio
A comparison, by division, of two pure numbers or two like denominate
numbers is a ratio.
The comparison by division of the pure numbers a and the like
8 miles
2 miles
denominate numbers are examples of ratios.
Rate
A comparison, by division, of two unlike denominate numbers is a rate.
The comparison by division of two unlike denominate numbers, such as
55 miles 40 dollars
1 gallon and 5 tickets
are examples of rates.
Let's agree to represent two numbers (pure or denominate) with the letters a
and b. This means that we're letting a represent some number and b
represent some, perhaps different, number. With this agreement, we can
write the ratio of the two numbers a and b as
or 7
a
|
The ratio “ is read as "a to b."
=|
The ratio — is read as" bto a."
a|o
Since a ratio or a rate can be expressed as a fraction, it may be reducible.
Sample Set B
Example:
The ratio 30 to 2 can be expressed as 2. Reducing, we get 2,
The ratio 30 to 2 is equivalent to the ratio 15 to 1.
Example:
The rate "4 televisions to 12 people" can be expressed as **levisions The
12 people
meaning of this rate is that "for every 4 televisions, there are 12 people."
: 1 television : : : "
Reducing, we get aRCuIcAe The meaning of this rate is that "for every 1
television, there are 3 people.”
Thus, the rate of "4 televisions to 12 people" is the same as the rate of "1
television to 3 people."
Practice Set B
Write the following ratios and rates as fractions.
Exercise:
Problem: 3 to 2
Solution:
bole
Exercise:
Problem: 1 to 9
Solution:
i
9
Exercise:
Problem: 5 books to 4 people
Solution:
5 books
4 people
Exercise:
Problem: 120 miles to 2 hours
Solution:
60 miles
1 hour
Exercise:
Problem: 8 liters to 3 liters
Solution:
8
3
Write the following ratios and rates in the form "a to 6." Reduce when
necessary.
Exercise:
Problem:
enfro
Solution:
9to5
Exercise:
col
Problem:
Solution:
1to3
Exercise:
25 miles
2 gallons
Problem:
Solution:
25 miles to 2 gallons
Exercise:
2 mechanics
Problem: 4 wrenches
Solution:
1 mechanic to 2 wrenches
Exercise:
. 15 video tapes
Problem: 18 video tapes
Solution:
5 to 6
Exercises
For the following 9 problems, complete the statements.
Exercise:
Problem:
Two numbers can be compared by subtraction if and only if .
Solution:
They are pure numbers or like denominate numbers.
Exercise:
Problem:
A comparison, by division, of two pure numbers or two like
denominate numbers is called a.
Exercise:
Problem:
A comparison, by division, of two unlike denominate numbers is
calleda.
Solution:
rate
Exercise:
Problem:
Exercise:
Problem:
Solution:
ratio
Exercise:
Problem:
Exercise:
Problem:
Solution:
rate
Exercise:
Problem:
Exercise:
Problem:
Solution:
ratio
11
b)
12
7 erasers
12 pencils
20 silver coins
35 gold coins
3 sprinklers
5 sprinklers
18 exhaust valves
11 exhaust valves
is an example of a. (ratio/rate)
is an example of a. (ratio/rate)
is an example of a. (ratio/rate)
is an example of a .(ratio/rate)
is an example of a. (ratio/rate)
is an example of a .(ratio/rate)
For the following 7 problems, write each ratio or rate as a verbal phrase.
Exercise:
Problem:
©2|00
Exercise:
Problem:
on|po
Solution:
two to five
Exercise:
8 feet
Problem: 3 seconds
Exercise:
29 miles
Problem: egallons
Solution:
29 mile per 2 gallons or 145 miles per 1 gallon
Exercise:
30,000 stars
Problem: 300 stars
Exercise:
5 yards
Problem: ian:
Solution:
510 2
Exercise:
164 trees
Problem: rE mee
For the following problems, write the simplified fractional form of each
ratio or rate.
Exercise:
Problem: 12 to 5
Solution:
12
5
Exercise:
Problem: 81 to 19
Exercise:
Problem: 42 plants to 5 homes
Solution:
42 plants
5 homes
Exercise:
Problem: 8 books to 7 desks
Exercise:
Problem: 16 pints to 1 quart
Solution:
16 pints
1 quart
Exercise:
Problem: 4 quarts to 1 gallon
Exercise:
Problem: 2.54 cm to 1 in
Solution:
2.54 cm
1 inch
Exercise:
Problem: 80 tables to 18 tables
Exercise:
Problem: 25 cars to 10 cars
Solution:
5
a
Exercise:
Problem: 37 wins to 16 losses
Exercise:
Problem: 105 hits to 315 at bats
Solution:
1 hit
3 at bats
Exercise:
Problem: 510 miles to 22 gallons
Exercise:
Problem: 1,042 characters to 1 page
Solution:
1,042 characters
1 page
Exercise:
Problem: 1,245 pages to 2 books
Exercises for Review
Exercise:
Problem: ({link]) Convert 2 to a mixed number.
Solution:
5
oo|e
Exercise:
Problem: ({link]) 1 2 of 22 is what number?
Exercise:
Problem: ({link]) Find the difference. ss — xz.
Solution:
299
1260
Exercise:
Problem:
({link]) Perform the division. If no repeating patterns seems to exist,
round the quotient to three decimal places: 22.35+17
Exercise:
Problem: ({link]) Find the value of 1.85 + 2 -4.1
Solution:
3.3875
Proportions
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses proportions. By the end of the
module students should be able to describe proportions and find the missing
factor in a proportion and be able to work with proportions involving rates.
Section Overview
e Ratios, Rates, and Proportions
e Finding the Missing Factor in a Proportion
e Proportions Involving Rates
Ratios, Rates, and Proportions
Ratio, Rate
We have defined a ratio as a comparison, by division, of two pure numbers
or two like denominate numbers. We have defined a rate as a comparison,
by division, of two unlike denominate numbers.
Proportion
A proportion is a statement that two ratios or rates are equal. The
following two examples show how to read proportions.
3 -. 8 25 miles = 50 miles
4 8 1 gallon 2 gallons
a ae | { {
3 is to 4 as 6 is to 8 25 miles is to 1 gallon as 50 miles is to 2 gallons
Sample Set A
Write or read each proportion.
Example:
oS ce tt)
Fe
3 is to 5. as 12 is to 20
Example:
10 items _ 2 items
5 dollars ~ 1 dollar
10 items is to 5 dollars as 2 items is to 1 dollar
Example:
8 is to 12 as 16 is to 24.
SB oh:
(Dimer
Example:
50 milligrams of vitamin C is to 1 tablet as 300 milligrams of vitamin C is
to 6 tablets.
50 _ 300
iat
Practice Set A
Write or read each proportion.
Exercise:
he eo Ey
Problem: 7 =
Solution:
3 is to 8 as 6 is to 16
Exercise:
2 1 10 1
Problem: —2P* — —Pept
1 window 5 windows
Solution:
2 people are to 1 window as 10 people are to 5 windows
Exercise:
Problem: 15 is to 4 as 75 is to 20.
Solution:
15 _ 75
4 20
Exercise:
Problem: 2 plates are to 1 tray as 20 plates are to 10 trays.
Solution:
2plates _ 20 plates
ltray 10 trays
Finding the Missing Factor in a Proportion
Many practical problems can be solved by writing the given information as
proportions. Such proportions will be composed of three specified numbers
and one unknown number. It is customary to let a letter, such as x, represent
the unknown number. An example of such a proportion is
z _ 20
4 ~~ 16
This proportion is read as " z is to 4 as 20 is to 16."
There is a method of solving these proportions that is based on the equality
of fractions. Recall that two fractions are equivalent if and only if their
cross products are equal. For example,
: 3 6
since y aa 5
3°8=6°4
24 = 24
=| co
DID
Notice that in a proportion that contains three specified numbers and a letter
representing an unknown quantity, that regardless of where the letter
appears, the following situation always occurs.
(number) - (letter) = (number) - (number)
————EEEEE ee
We recognize this as a multiplication statement. Specifically, it is a missing
factor statement. (See [link] for a discussion of multiplication statements. )
For example,
f= if means that 16-x2 = 4- 20
-=3 means that 4-20 =— 16-2
ae means that 5-16—4-z2z
5
4
= a means that 5-2 = 4- 20
Each of these statements is a multiplication statement. Specifically, each is
a missing factor statement. (The letter used here is z, whereas M was used
in [link].)
Finding the Missing Factor in a Proportion
The missing factor in a missing factor statement can be determined by
dividing the product by the known factor, that is, if « represents the missing
factor, then
x = (product) + (known factor)
Sample Set B
Find the unknown number in each proportion.
Example:
i= 2. Find the cross product.
Ge ae 20:
16572 =" 80 Divide the product 80 by the known factor 16.
80
16
49 — ns The unknown number is 5.
This mean that A = a or 5 is to 4 as 20 is to 16.
4G =
Example:
5
5 = a. Find the cross product.
Ge eax
80 = 20-2 Divide the product 80 by the known factor 20.
v=
20
4 =r The unknown number is 4.
This means that = 2 or, 5 is to 4 as 20 is to 6.
Example:
— = as Find the cross product.
16-2 = 64-3
16s — 192" Divide:l97 by. 16:
192
LE aes
L =e? The unknown number is 12.
The means that -- = or, 16 is to 3 as 64 is to 12.
Example:
- = ani Find the cross products.
9-AQ° = 8 x
360 = 8-2 Divide 360 by 8.
M = g
45 a6 The unknown number is 45.
Practice Set B
Find the unknown number in each proportion.
Exercise:
Problem: + = 2
Solution:
r=3
Exercise:
Problem: a. = M4
Solution:
x=5
Exercise:
Problem: ia =£
Solution:
x = 45
Exercise:
Problem: + = =
Solution:
= 48
8
6
Proportions Involving Rates
Recall that a rate is a comparison, by division, of unlike denominate
numbers. We must be careful when setting up proportions that involve rates.
The form is important. For example, if a rate involves two types of units,
Say unit type 1 and unit type 2, we can write
unit type 1 =
unit type 2
or
unit type 1 _
unit type 1
Same units
appear on
same side.
Same units appear
unit type jer on same side.
unit type 2 <- Same units appear
on same side.
unit type 2
unit type 2
t
Same units
appear on
same side.
Both cross products produce a statement of the type
(unit type 1) - (unit type 2) = (unit type 1) - (unit type 2)
which we take to mean the comparison
(unit type 1) is to (unit type 2) as (unit type 1) is to (unit type 2)
Comparison of type 1 Comparison of type 1
with type 2 with type 2
Same overall type
Examples of correctly expressed proportions are the following:
FIE
N
5
z
:
§
F
&
&
However, if we write the same type of units on different sides, such as,
unit typel —— unit type 2
unit type 2 unit type 1
the cross product produces a statement of the form
(unit type 1) - (unit type 1) = (unit type 2) - (unit type 2)
Comparison of type 1 Comparison of type 2
with type 1 with type 2
Different overall types
We can see that this is an incorrect comparison by observing the following
example: It is incorrect to write
2hooks __ 6 poles
3poles ~ 4 hooks
for two reason.
1. The cross product is numerically wrong: (2-4 4 3-6).
2. The cross product produces the statement “hooks are to hooks as poles
are to poles,” which makes no sense.
Exercises
Exercise:
Problem: A statement that two ratios or are equal is calleda.
Solution:
rates, proportion
For the following 9 problems, write each proportion in fractional form.
Exercise:
Problem: 3 is to 7 as 18 is to 42.
Exercise:
Problem: 1 is to 11 as 3 is to 33.
Solution:
Exercise:
Problem: 9 is to 14 as 27 is to 42.
Exercise:
Problem: 6 is to 90 as 3 is to 45.
Solution:
Exercise:
Problem: 5 liters is to 1 bottle as 20 liters is to 4 bottles.
Exercise:
Problem:
18 grams of cobalt is to 10 grams of silver as 36 grams of cobalt is to
20 grams of silver.
Solution:
18 grcobalt — 36 grcobalt
10 gr silver 20 gr silver
Exercise:
Problem:
4 cups of water is to 1 cup of sugar as 32 cups of water is to 8 cups of
sugar.
Exercise:
Problem:
3 people absent is to 31 people present as 15 people absent is to 155
people present.
Solution:
3 people absent _— 15 people absent
31 people present ‘155 people present
Exercise:
Problem: 6 dollars is to 1 hour as 90 dollars is to 15 hours.
For the following 10 problems, write each proportion as a sentence.
Exercise:
ao eee! Wie
Problem: Fi 5p
Solution:
3 is to 4as 15 is to 20
Exercise:
otk. a 438
Problem: 7 =
Exercise:
. sjoggers _ 6 joggers
Problem: 100 feet ~ 200 feet
Solution:
3 joggers are to 100 feet as 6 joggers are to 200 feet
Exercise:
. 12 marshmallows __ 36 marshmallows
Problem: 3 sticks _ 9 sticks
Exercise:
. 40 miles _ 2 gallons
Problem: 80 miles —« 4 gallons
Solution:
AO miles are to 80 miles as 2 gallons are to 4 gallons
Exercise:
. 4couches __ 2 houses
Problem: 10 couches ~— 5 houses
Exercise:
. Lperson _ 8 people
Problem: ——,,— sob =e bey
Solution:
1 person is to 1 job as 8 people are to 8 jobs
Exercise:
1 popsicle 5 popsicle
Problem: > Giidren = ~1child
Exercise:
, 2,000 pounds _ 60,000 pounds
Problem: 1 ton = 30 tons
Solution:
2,000 pounds are to 1 ton as 60,000 pounds are to 30 tons
Exercise:
1 table __ 2 people
Problem: 5tables ~— 10 people
For the following 10 problems, solve each proportion.
Exercise:
6 G
Problem: - = 5
Solution:
Heat
Exercise:
Cte a 28
Problem: 0 =
Exercise:
Problem: 2 — 22
x 16
Solution:
58
Exercise:
Problem: — = =
Exercise:
Problem:
Solution:
P= 5
Exercise:
Problem: ——
Exercise:
Problem:
Solution:
oA
Exercise:
Problem: —
Exercise:
Problem:
Solution:
o= AQ
Exercise:
© |00
P's eee
Problem: 30 = GO
For the following 5 problems, express each sentence as a proportion then
solve the proportion.
Exercise:
Problem: 5 hats are to 4 coats as z hats are to 24 coats.
Solution:
p= 30
Exercise:
Problem: x cushions are to 2 sofas as 24 cushions are to 16 sofas.
Exercise:
Problem:
1 spacecraft is to 7 astronauts as 5 spacecraft are to x astronauts.
Solution:
x= 39d
Exercise:
Problem:
56 microchips are to x circuit boards as 168 microchips are to 3 circuit
boards.
Exercise:
Problem:
18 calculators are to 90 calculators as x students are to 150 students.
Solution:
x= 30
Exercise:
Problem: z dollars are to $40,000 as 2 sacks are to 1 sack.
Indicate whether the proportion is true or false.
Exercise:
Problem: 16 — 64
Solution:
true
Exercise:
Problem: p= we
Exercise:
ao
Problem: 9 > 36
Solution:
false
Exercise:
. Sknives __ 12 knives
Problem: 7forks ~ 15 forks
Exercise:
33 miles __ 99 miles
Problem: lgallon —- 3 gallons
Solution:
true
Exercise:
. 320feet _ 65 feet
Problem: 5 seconds 1 second
Exercise:
. d5students _ lclass
Problem: 70 students ~——- 2 classes
Solution:
true
Exercise:
. 9mlchloride _ 1 test tube
Problem: 45 ml chloride ~—-7_:~ test tubes
Exercises for Review
Exercise:
Problem:
({link]) Use the number 5 and 7 to illustrate the commutative property
of addition.
Solution:
5+7=—12
7+5-—12
Exercise:
Problem:
({link]) Use the numbers 5 and 7 to illustrate the commutative property
of multiplication.
Exercise:
Problem: ({link]) Find the difference. a = +.
Solution:
ava
ii
Exercise:
Problem: ({link]) Find the product. 8.06129 - 1,000.
Exercise:
Problem:
({link]) Write the simplified fractional form of the rate “sixteen
sentences to two paragraphs.”
Solution:
8 sentences
1 paragraph
Applications of Proportions
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses applications of proportions. By
the end of the module students should be able to solve proportion problems
using the five-step method.
Section Overview
e The Five-Step Method
e Problem Solving
The Five-Step Method
In [link] we noted that many practical problems can be solved by writing
the given information as proportions. Such proportions will be composed of
three specified numbers and one unknown number represented by a letter.
The first and most important part of solving a proportion problem is to
determine, by careful reading, what the unknown quantity is and to
represent it with some letter.
The Five-Step Method
The five-step method for solving proportion problems:
1. By careful reading, determine what the unknown quantity is and
represent it with some letter. There will be only one unknown in a
problem.
2. Identify the three specified numbers.
3. Determine which comparisons are to be made and set up the
proportion.
4. Solve the proportion (using the methods of [link]).
5. Interpret and write a conclusion in a sentence with the appropriate
units of measure.
Step 1 is extremely important. Many problems go unsolved because time is
not taken to establish what quantity is to be found.
When solving an applied problem, always begin by determining the
unknown quantity and representing it with a letter.
Problem Solving
Sample Set A
Example:
On a map, 2 inches represents 25 miles. How many miles are represented
by 8 inches?
Step 1 The unknown quantity is miles.
Let x = number of miles represented by 8 inches
Step 2 The three specified numbers are
2 inches
25 miles
8 inches
Step 3 The comparisons are
. : 2 inches
2 inches to 25 miles — 25 miles
8 inches
x miles
Proportions involving ratios and rates are more readily solved by
suspending the units while doing the computations.
2 8
8 inches to x miles >
aise 4 cz = “ Perform the cross multiplication.
PAS te) 8 PA
2-x = 200 Divide 200 by 2.
2 =
c= 100
In step 1, we let x represent the number of miles. So, z represents 100
miles.
Step 5 If 2 inches represents 25 miles, then 8 inches represents 100
miles.
Try [Link] in [linkl.
Example:
An acid solution is composed of 7 parts water to 2 parts acid. How many
parts of water are there in a solution composed of 20 parts acid?
e Step 1 The unknown quantity is the number of parts of water.
Let n = number of parts of water.
e Step 2 The three specified numbers are
7 parts water
2 parts acid
20 parts acid
e Step 3 The comparisons are
7 parts water to 2 parts acid > =
nr
n parts water to 20 parts acid > 35
a
7 n
me 20!
e Step 4 5 = 3 Perform the cross multiplication.
fie AU Ae)
140 = 2-n Divide 140 by 2.
70
In step 1 we let n represent the number of parts of water. So, n
represents 70 parts of water.
e Step 5 7 parts water to 2 parts acid indicates 70 parts water to 20 parts
acid.
Try [link] in [link].
Example:
A 5-foot girl casts a 3+--foot shadow at a particular time of the day. How
tall is a person who casts a 3-foot shadow at the same time of the day?
e Step 1 The unknown quantity is the height of the person.
Let h = height of the person.
e Step 2 The three specified numbers are
5 feet ( height of girl)
35 feet (length of shadow)
3 feet (length of shadow)
e Step 3 The comparisons are
5-foot girl is to 3 foot shadow > ae
h-foot person is to 3-foot shadow — h
a=}
° Step 4 31 = 4
bed See
15 = ~-h Dividel5by 2
3 = Ah
3
Tah
2
2 =h
h = 45
e Step 5 A person who casts a 3-foot shadow at this particular time of
the day is 45 feet tall.
Try [Link] in [link].
Example:
The ratio of men to women in a particular town is 3 to 5. How many
women are there in the town if there are 19,200 men in town?
e Step 1 The unknown quantity is the number of women in town.
Let x = number of women in town.
e Step 2 The three specified numbers are
3)
5
19,200
e Step 3 The comparisons are 3 men to 5 women > 2
5
192
19,200 men to x women > 18,200
3 _ 19,200
; Sena _ 19,200
ao 19000 5
3-2 — 96,000
96,000
3
Ls 2 VOU
e Step 5 There are 32,000 women in town.
Example:
The rate of wins to losses of a particular baseball team is 5 . How many
games did this team lose if they won 63 games?
e Step 1 The unknown quantity is the number of games lost.
Let n = number of games lost.
e Step 2 Since 3 — means 9 wins to 2 losses, the three specified
numbers are
9 (wins)
2 (losses)
63 (wins)
e Step 3 The comparisons are
9 wins to 2 losses > +
63 wins to n losses > “
Gere eGo:
2 n
° Step 4 2 = on
ot — 22 63
97 = 126
n=
i — we
e Step 5: This team had 14 losses.
Try [link] in [ink].
Practice Set A
Solve each problem.
Exercise:
Problem:
On a map, 3 inches represents 100 miles. How many miles are
represented by 15 inches?
e Step 1
e Step 2
e Step 3
e Step 4
e Step5
Solution:
500 miles
Exercise:
Problem:
An alcohol solution is composed of 14 parts water to 3 parts alcohol.
How many parts of alcohol are in a solution that is composed of 112
parts water?
e Step 1
e Step 2
e Step 3
e Step 4
e Step5
Solution:
24 parts of alcohol
Exercise:
Problem:
A 5+ -foot woman casts a 7-foot shadow at a particular time of the
day. How long of a shadow does a 3-foot boy cast at that same time of
day?
e Step 1
e Step 2
e Step 3
e Step 4
e Step5
Solution:
9
2 ar feet
Exercise:
Problem:
The rate of houseplants to outside plants at a nursery is 4 to 9. If there
are 384 houseplants in the nursery, how many outside plants are there?
e Step 1
e Step 2
e Step 3
e Step 4
e Step5
Solution:
864 outside plants
Exercise:
Problem:
The odds for a particular event occurring are 11 to 2. (For every 11
times the event does occur, it will not occur 2 times.) How many times
does the event occur if it does not occur 18 times?
e Step 1
e Step 2
e Step 3
e Step 4
e Step 5
Solution:
The event occurs 99 times.
Exercise:
Problem:
The rate of passing grades to failing grades in a particular chemistry
class is £ . If there are 21 passing grades, how many failing grades are
there?
e Step 1
e Step 2
e Step 3
e Step 4
e Step5
Solution:
6 failing grades
Exercises
For the following 20 problems, use the five-step method to solve each
problem.
Exercise:
Problem:
On a map, 4 inches represents 50 miles. How many inches represent
300 miles?
Solution:
24
Exercise:
Problem:
On a blueprint for a house, 2 inches represents 3 feet. How many
inches represent 10 feet?
Exercise:
Problem:
A model is built to * scale. If a particular part of the model measures
6 inches, how long is the actual structure?
Solution:
45 inches
Exercise:
Problem:
An acid solution is composed of 5 parts acid to 9 parts of water. How
many parts of acid are there in a solution that contains 108 parts of
water?
Exercise:
Problem:
An alloy contains 3 parts of nickel to 4 parts of silver. How much
nickel is in an alloy that contains 44 parts of silver?
Solution:
33 parts
Exercise:
Problem:
The ratio of water to salt in a test tube is 5 to 2. How much salt is ina
test tube that contains 35 ml of water?
Exercise:
Problem:
The ratio of sulfur to air in a container is =: How many ml of air are
there in a container that contains 207 ml of sulfur?
Solution:
2328.75
Exercise:
Problem:
A 6-foot man casts a 4-foot shadow at a particular time of the day.
How tall is a person that casts a 3-foot shadow at that same time of the
day?
Exercise:
Problem:
A 5 +-foot woman casts a 1 5 foot shadow at a particular time of the
day. How long a shadow does her 3 +-foot niece cast at the same time
of the day?
Solution:
21
oy feet
Exercise:
Problem:
A man, who is 6 feet tall, casts a 7-foot shadow at a particular time of
the day. How tall is a tree that casts an 84-foot shadow at that same
time of the day?
Exercise:
Problem:
The ratio of books to shelves in a bookstore is 350 to 3. How many
books are there in a store that has 105 shelves?
Solution:
12,250
Exercise:
Problem:
The ratio of algebra classes to geometry classes at a particular
community college is 13 to 2. How many geometry classes does this
college offer if it offers 13 algebra classes?
Exercise:
Problem:
The odds for a particular event to occur are 16 to 3. If this event occurs
64 times, how many times would you predict it does not occur?
Solution:
12
Exercise:
Problem:
The odds against a particular event occurring are 8 to 3. If this event
does occur 64 times, how many times would you predict it does not
occur?
Exercise:
Problem:
The owner of a stationery store knows that a 1-inch stack of paper
contains 300 sheets. The owner wishes to stack the paper in units of
550 sheets. How many inches tall should each stack be?
Solution:
1
lon
Exercise:
Problem:
A recipe that requires 6 cups of sugar for 15 servings is to be used to
make 45 servings. How much sugar will be needed?
Exercise:
Problem:
A pond loses 7s gallons of water every 2 days due to evaporation.
How many gallons of water are lost, due to evaporation, in 5 day?
Solution:
i
oo|~I
Exercise:
Problem:
A photograph that measures 3 inches wide and 4s inches high is to be
enlarged so that it is 5 inches wide. How high will it be?
Exercise:
Problem:
If 25 pounds of fertilizer covers 400 square feet of grass, how many
pounds will it take to cover 500 square feet of grass?
Solution:
1
31 =
Exercise:
Problem:
Every 15 teaspoons of a particular multiple vitamin, in granular form,
contains 0.65 the minimum daily requirement of vitamin C. How many
teaspoons of this vitamin are required to supply 1.25 the minimum
daily requirement?
Exercises for Review
Exercise:
Problem: ({link]) Find the product, 818 - 0.
Solution:
0
Exercise:
Problem: ([link]) Determine the missing numerator: - = ae
Exercise:
3 4
to ta
19
1
20
Problem: ([link]) Find the value of
Solution:
ewo|bo
Exercise:
Problem: ({link]) Subtract 0.249 from the sum of 0.344 and 0.612.
Exercise:
(i “4. 6 _ 36
Problem: ([link]) Solve the proportion: ~ = 3.
Solution:
5
Percent
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses percents. By the end of the
module students should understand the relationship between ratios and
percents and be able to make conversions between fractions, decimals, and
percents.
Section Overview
e Ratios and Percents
e The Relationship Between Fractions, Decimals, and Percents —
Making Conversions
Ratios and Percents
Ratio, Percent
We defined a ratio as a comparison, by division, of two pure numbers or
two like denominate numbers. A most convenient number to compare
numbers to is 100. Ratios in which one number is compared to 100 are
called percents. The word percent comes from the Latin word "per
centum." The word "per" means "for each" or "for every," and the word
"centum" means "hundred." Thus, we have the following definition.
Percent means “for each hundred," or "for every hundred."
The symbol % is used to represent the word percent.
Sample Set A
Example:
The ratio 26 to 100 can be written as 26%. We read 26% as "twenty-six
percent."
Example:
The ratio ee. can be written as 165%.
We read 165% as "one hundred sixty-five percent."
Example:
The percent 38% can be written as the fraction s.
Example:
The percent 210% can be written as the fraction par or the mixed number
10
2 00 OY eae LP
Example:
Since one dollar is 100 cents, 25 cents is —- of a dollar. This implies that
25 cents is 25% of one dollar.
Practice Set A
Exercise:
Problem: Write the ratio 16 to 100 as a percent.
Solution:
16%
Exercise:
Problem: Write the ratio 195 to 100 as a percent.
Solution:
195%
Exercise:
Problem: Write the percent 83% as a ratio in fractional form.
Solution:
83 _
100
Exercise:
Problem: Write the percent 362% as a ratio in fractional form.
Solution:
362 181
100 CF 50
The Relationship Between Fractions, Decimals, and Percents —
Making Conversions
Since a percent is a ratio, and a ratio can be written as a fraction, and a
fraction can be written as a decimal, any of these forms can be converted to
any other.
Before we proceed to the problems in [link] and [link], let's summarize the
conversion techniques.
To Convert a Fraction To Convert a To Convert a
Decimal Percent
To a fraction: To a decimal:
To a decimal: Divide the Beagine Move the —
numerator by the decimal and decimal point 2
denominator reduce the places to the left
resulting and drop the %
fraction symbol
To a percent: Convert the To a percent: To a fraction:
nae
fraction first to a decimal, Move the Drop the % sign
then move the decimal decimal point 2 and write the
point 2 places to the right places se the number over"
and affix the % symbol right and affix 100. Reduce, if
the % symbol possible.
Conversion Techniques — Fractions, Decimals, Percents
Sample Set B
Example:
Convert 12% to a decimal.
oa
2 — io = 0.12
Note that
12% = 12.% = 0.12
Lest
The % symbol is dropped, and the decimal point moves 2 places to the left.
Example:
Convert 0.75 to a percent.
Seon
0.75 = B= 75%
Note that
0.75 = 75% = 75.%
_—_——
The % symbol is affixed, and the decimal point moves 2 units to the right.
Example:
Convert 2. to a percent.
We see in [link] that we can convert a decimal to a percent. We also know
that we can convert a fraction to a decimal. Thus, we can see that if we first
convert the fraction to a decimal, we can then convert the decimal to a
percent.
or? =06= = 4 =60%
3
5 100
on
ww W
COP!lD
Example:
Convert 42% to a fraction.
— 42 _ 21
42% = 100 ~ 50
or
42% =0.42= 2 =H
Practice Set B
Exercise:
Problem: Convert 21% to a decimal.
Solution:
0.21
Exercise:
Problem: Convert 461% to a decimal.
Solution:
4.61
Exercise:
Problem: Convert 0.55 to a percent.
Solution:
55%
Exercise:
Problem: Convert 5.64 to a percent.
Solution:
564%
Exercise:
Problem: Convert on to a percent.
Solution:
15%
Exercise:
Problem: Convert + to a percent
Solution:
137.5%
Exercise:
Problem: Convert “. to a percent.
Solution:
27.21%
Exercises
For the following 12 problems, convert each decimal to a percent.
Exercise:
Problem: 0.25
Solution:
25%
Exercise:
Problem: 0.36
Exercise:
Problem: 0.48
Solution:
48%
Exercise:
Problem: 0.343
Exercise:
Problem: 0.771
Solution:
77.1%
Exercise:
Problem: 1.42
Exercise:
Problem: 2.58
Solution:
258%
Exercise:
Problem: 4.976
Exercise:
Problem: 16.1814
Solution:
1,618.14%
Exercise:
Problem: 533.01
Exercise:
Problem: 2
Solution:
200%
Exercise:
Problem: 14
For the following 10 problems, convert each percent to a decimal.
Exercise:
Problem: 15%
Solution:
0.15
Exercise:
Problem: 43%
Exercise:
Problem: 16.2%
Solution:
0.162
Exercise:
Problem
Exercise:
Problem
200.070
25.0570
Solution:
0.0505
Exercise:
Problem
Exercise:
Problem
6.11%
: 0.78%
Solution:
0.0078
Exercise:
Problem
Exercise:
Problem
: 0.88%
: 0.09%
Solution:
0.0009
Exercise:
Problem
: 0.001%
For the following 14 problems, convert each fraction to a percent.
Exercise:
Problem:
one
Solution:
20%
Exercise:
Problem:
onloo
Exercise:
Colon
Problem:
Solution:
62.5%
Exercise:
Problem:
Exercise:
Problem:
Solution:
28%
Exercise:
Problem: —
Exercise:
Problem:
Solution:
49.09%
Exercise:
Problem:
Exercise:
Problem:
Solution:
164%
Exercise:
Problem:
Exercise:
Problem:
Solution:
945%
Exercise:
27
55
41
25
Problem: =~
Exercise:
. 6
Problem: aT
Solution:
54.54%
Exercise:
. 35
Problem: 7
For the following 14 problems, convert each percent to a fraction.
Exercise:
Problem: 80%
Solution:
4
5
Exercise:
Problem: 60%
Exercise:
Problem: 25%
Solution:
1
4
Exercise:
Problem: 75%
Exercise:
Problem: 65%
Solution:
18:
20
Exercise:
Problem: 18%
Exercise:
Problem: 12.5%
Solution:
4)
8
Exercise:
Problem: 37.5%
Exercise:
Problem: 512.5%
Solution:
Al
1
3 or a5
Exercise:
Problem: 937.5%
Exercise:
Problem: 9.9%
Solution:
=
10
Exercise:
Problem: 55.5%
Exercise:
Problem: 22. 2%
Solution:
2
9
Exercise:
Problem: 63.6%
Exercises for Review
Exercise:
oni ; ; 40 . 97
Problem: ((link]) Find the quotient. <7; + 857.
Solution:
4
45
Exercise:
Problem: ({link}) 4 of what number is 22?
Exercise:
«71 : 25 yy ft 8.
Problem: ({link]) Find the value of er ag oe
Solution:
129 9 93.
“¢0 OF 2% = 239
Exercise:
Problem: ({link]) Round 6.99997 to the nearest ten thousandths.
Exercise:
Problem:
({link]) On a map, 3 inches represent 40 miles. How many inches
represent 480 miles?
Solution:
36 inches
Fractions of One Percent
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses fractions of one percent. By the
end of the module students should understand the meaning of a fraction of
one percent and be able to make conversions involving fractions of one
percent.
Section Overview
¢ Conversions Involving Fractions of One Percent
¢ Conversions Involving Nonterminating Fractions
Conversions Involving Fractions of One Percent
Percents such as 5%, 2 %, 3 %, and = %, where 1% has not been
attained, are fractions of 1%. This implies that
5% = 5 of 1%
2% = 2 of 1%
2% = 3 of 1%
EH%= Ff of 1%
Since "percent" means "for each hundred," and "of" means "times," we
have
(6. 4 en ee eee |
77 = 7b 1% = 2° Oy = 200
307 _ 3 re eh} oh, hee,
3% = 508 1% = = 100 ~ 500
507 _ 5 pie BNA oo
3% = 308 1% = 3 100 ~
To 7 53 oa TU oe
qi = 77of 1h = FG 100 ~
Sample Set A
Example:
Convert 2% to a fraction.
2% = 20f1% = 2.1,
3
50
ey velely
= 2
=) ok.
= 150
Example:
Convert 2 % to a decimal.
3% = $ofl% = 3. ay
=: 0.625-0.01
0.00625
Practice Set A
Exercise:
Problem: Convert + % to a fraction.
Solution:
1
400
Exercise:
Problem: Convert ~ % to a fraction.
Solution:
3
800
Exercise:
Problem: Convert 35 % to a fraction.
Solution:
et
30
Conversions Involving Nonterminating Fractions
We must be careful when changing a fraction of 1% to a decimal. The
number 2. as we know, has a nonterminating decimal representation.
Therefore, it cannot be expressed exactly as a decimal.
When converting nonterminating fractions of 1% to decimals, it is
customary to express the fraction as a rounded decimal with at least three
decimal places.
Converting a Nonterminating Fraction to a Decimal
To convert a nonterminating fraction of 1% to a decimal:
1. Convert the fraction as a rounded decimal.
2. Move the decimal point two digits to the left and remove the percent
sign.
Sample Set B
Example:
Convert £ % to a three-place decimal.
1. Convert - to a decimal.
Since we wish the resulting decimal to have three decimal digits, and
removing the percent sign will account for two of them, we need to
round = to one place (2 + 1 = 3).
2% = 0.7% to one decimal place. (2 = 0.6666...)
2. Move the decimal point two digits to the left and remove the % sign.
We'll need to add zeros to locate the decimal point in the correct
location.
2% = 0.007 to 3 decimal places
Example:
Convert 5% to a four-place decimal.
1. Since we wish the resulting decimal to have four decimal places, and
removing the percent sign will account for two, we to round “ to
two places.
5% = 5.36% to two decimal places. (=+ = 0.3636...)
2. Move the decimal point two places to the left and drop the percent
sign.
5% = 0.0536 to four decimal places.
Example:
Convert 28 a % to a decimal rounded to ten thousandths.
1. Since we wish the resulting decimal to be rounded to ten thousandths
(four decimal places), and removing the percent sign will account for
two, we need to round = to two places.
282% = 28.56% to two decimal places. (2 = 0.5555...)
2. Move the decimal point to the left two places and drop the percent
sign.
282% — 0.2856 correct to ten thousandths.
Practice Set B
Exercise:
Problem: Convert £% to a three-place decimal.
Solution:
0.008
Exercise:
Problem: Convert 51 2% to a decimal rounded to ten thousandths.
Solution:
0.5145
Exercises
Make the conversions as indicated.
Exercise:
Problem: Convert “ % to a fraction.
Solution:
ao
400
Exercise:
Problem: Convert 2% to a fraction.
Exercise:
Problem: Convert 5% to a fraction.
Solution:
=
900
Exercise:
Problem: Convert % to a fraction.
Exercise:
Problem
: Convert > % to a fraction.
Solution:
i eee
400 OF 80
Exercise:
Problem
Exercise:
Problem
: Convert £% to a fraction.
: Convert 1 2% to a fraction.
Solution:
dae
700
Exercise:
Problem: Convert 2 7% to a fraction.
Exercise:
Problem: Convert 254% to a fraction.
Solution:
101
400
Exercise:
Problem: Convert 50% to a fraction.
Exercise:
Problem: Convert 722% to a fraction.
Solution:
363
500
Exercise:
Problem: Convert 995% to a fraction.
Exercise:
Problem: Convert 1364% to a fraction.
Solution:
AL
30
Exercise:
Problem: Convert 5214% to a fraction.
Exercise:
Problem: Convert 102% to a decimal.
Solution:
OLS ss
2 = 0.102
Exercise:
Problem: Convert 122% to a decimal.
Exercise:
Problem: Convert 3 5% to a decimal.
Solution:
31
3 = 0.03875
Exercise:
Problem: Convert 7% to a decimal.
Exercise:
Problem: Convert 2% to a three-place decimal.
Solution:
0.004
Exercise:
Problem: Convert 5% to a three-place decimal.
Exercise:
Problem: Convert 6 3% to a four-place decimal.
Solution:
0.0627
Exercise:
Problem: Convert 9 2% to a four-place decimal.
Exercise:
Problem: Convert 242-% to a three-place decimal.
Solution:
0.242
Exercise:
Problem: Convert 45-5-% to a three-place decimal.
Exercise:
Problem: Convert 11-3 % to a four-place decimal.
Solution:
0.1194
Exercise:
Problem: Convert 5 7% to a three-place decimal.
Exercises for Review
Exercise:
Problem
: (Llink]) Write 8-8-8-8-8 using exponents.
Solution:
g5
Exercise:
Problem
Exercise:
Problem
: (Llink]) Convert 4t to an improper fraction.
ae 7 2 1
: (Llink]) Find the sum. jo tart 7
Solution:
197
210
Exercise:
Problem
Exercise:
Problem
: ([link]) Find the product. (4.21)(0.006).
: (Llink]) Convert 8.062 to a percent.
Solution:
806.2%
Applications of Percents
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module
discusses applications of percents. By the end of the module students should be able to distinguish between
base, percent, and percentage and be able to find the percentage, the percent, and the base.
Section Overview
e Base, Percent, and Percentage
e Finding the Percentage
e Finding the Percent
e Finding the Base
Base, Percent, and Percentage
There are three basic types of percent problems. Each type involves a base, a percent, and a percentage, and
when they are translated from words to mathematical symbols each becomes a multiplication statement.
Examples of these types of problems are the following:
1. What number is 30% of 50? (Missing product statement.)
2. 15 is what percent of 50? (Missing factor statement.)
3. 15 is 30% of what number? (Missing factor statement.)
In problem 1, the product is missing. To solve the problem, we represent the missing product with P.
P = 30% - 50
Percentage
The missing product P is called the percentage. Percentage means part, or portion. In P = 30% - 50, P
represents a particular part of 50.
In problem 2, one of the factors is missing. Here we represent the missing factor with Q.
15=Q-50
Percent
The missing factor is the percent. Percent, we know, means per 100, or part of 100. In 15 = Q - 50, Q
indicates what part of 50 is being taken or considered. Specifically, 15 = Q - 50 means that if 50 was to be
divided into 100 equal parts, then Q indicates 15 are being considered.
In problem 3, one of the factors is missing. Represent the missing factor with B.
15=30%-B
Base
The missing factor is the base. Some meanings of base are a source of supply, or a starting place. In
15 = 30% - B, B indicates the amount of supply. Specifically, 15 = 30% - B indicates that 15 represents
30% of the total supply.
Each of these three types of problems is of the form
(percentage) = (percent) - (base)
We can determine any one of the three values given the other two using the methods discussed in [link].
Finding the Percentage
Sample Set A
Example:
What number is 30% of 50? Missing product statement.
aa ey d 1 1
(percentage) = (percent) - (base)
A 4 4 - 4
P = 30% : 50 Convert 30% to a decimal.
P = .30 : 50 Multiply.
IP = 15
Thus, 15 is 30% of 50.
Do [link], [link].
Example:
What number is 36% of 95? Missing product statement.
oe L 1 1
(percentage) = (percent) - (base)
A 4 4 2
P = 36% : 95 Convert 36% to a decimal.
P = .36 : 95 Multiply
IP = 34.2
Thus, 34.2 is 36% of 95.
Do [link!, [link].
Example:
A salesperson, who gets a commission of 12% of each sale she makes, makes a sale of $8,400.00. How
much is her commission?
We need to determine what part of $8,400.00 is to be taken. What part indicates percentage.
What number is 12% of 8,400.00? Missing product statement.
(percentage) = (percent) - (base)
t 4 i . i
1p = 12% - 8,400.00 Convert to decimals.
P = ip - 8,400.00 Multiply.
JP: = 1008.00
Thus, the salesperson's commission is $1,008.00.
Do [link], [link].
Example:
A girl, by practicing typing on her home computer, has been able to increase her typing speed by 110%. If
she originally typed 16 words per minute, by how many words per minute was she able to increase her
speed?
We need to determine what part of 16 has been taken. What part indicates percentage.
What number is 110% of 16? Missing product statement.
So Oe L 1 t
(percentage) = (percent) - (base)
t 4 ’ oa
ie =) 10% : 16 Convert to decimals.
P = 1.10 : 16 =Multiply.
IP = 17.6
Thus, the girl has increased her typing speed by 17.6 words per minute. Her new speed is 16 + 17.6 = 33.6
words per minute.
Do [link], [link].
Example:
A student who makes $125 a month working part-time receives a 4% salary raise. What is the student's new
monthly salary?
With a 4% raise, this student will make 100% of the original salary + 4% of the original salary. This means
the new salary will be 104% of the original salary. We need to determine what part of $125 is to be taken.
What part indicates percentage.
What number is 104% of 125 Missing product statement.
<> I + 1 +
(percentage) = (percent) - (base)
4 + 4 4
ie = 104% - 125 Convert to decimals.
Ip = 1.04 . 125 Multiply.
IZ = 130
Thus, this student's new monthly salary is $130.
Do [link], [link].
Example:
An article of clothing is on sale at 15% off the marked price. If the marked price is $24.95, what is the sale
price?
Since the item is discounted 15%, the new price will be 100% — 15% = 85% of the marked price. We need
to determine what part of 24.95 is to be taken. What part indicates percentage.
What number is 85% of $24.95. Missing product statement.
—_— + 1 +
(percentage) = (percent) - (base)
4 4 4 + 4
P = 85% - 24.95 Convert to decimals.
IP = 85 - 24.95 Multiply.
iP 212075 Since this number represents money,
we'll round to 2 decimal places
JP = 21.21
Thus, the sale price of the item is $21.21.
Practice Set A
Exercise:
Problem: What number is 42% of 85?
Solution:
33:7
Exercise:
Problem:
A sales person makes a commission of 16% on each sale he makes. How much is his commission if he
makes a sale of $8,500?
Solution:
$1,360
Exercise:
Problem:
An assembly line worker can assemble 14 parts of a product in one hour. If he can increase his
assembly speed by 35%, by how many parts per hour would he increase his assembly of products?
Solution:
4.9
Exercise:
Problem:
A computer scientist in the Silicon Valley makes $42,000 annually. What would this scientist's new
annual salary be if she were to receive an 8% raise?
Solution:
$45,360
Finding the Percent
Sample Set B
Example:
15 is what percent of 50? Missing factor statement.
4 {—_—~"_—s«W6gSt- 4
(percentage) = (percent) i (base) [(product) = (factor) - (factor)]
+ + 1 1
15 = : 50
Recall that (missing factor) = (product) + (known factor).
Q = 15+50 Divide.
Q = 0.3 Convert to a percent.
Oo = 30%
Thus, 15 is 30% of 50.
Do [link], [link].
Example:
4.32 i: what percent e dee Missing factor statement.
Se
(percentage) = (percent) ; (base) [(product) = (factor) - (factor)]
+ + 1 4
4.32 = Q : 72
Q = 4.32+72 Divide.
Qe 0.06 Convert to a percent.
= 6%
Thus, 4.32 is 6% of 72.
Do [link], [link].
Example:
On a 160 question exam, a student got 125 correct answers. What percent is this? Round the result to two
decimal places.
We need to determine the percent.
aie i what percent o poe Missing factor statement.
ae
(percentage) = (percent) ; (base) [(product) = (factor) - (factor)]
+ + 4 4
125 = Q - 160
Q = 125+160 Divide.
Q 0.78125 Round to two decimal places.
Q = ww
Thus, this student received a 78% on the exam.
Do [link], [link].
Example:
A bottle contains 80 milliliters of hydrochloric acid (HCl) and 30 milliliters of water. What percent of HCl
does the bottle contain? Round the result to two decimal places.
We need to determine the percent. The total amount of liquid in the bottle is
80 milliliters + 30 milliliters = 110 milliliters.
ae i: what percent o pte Missing factor statement.
SN
(percentage) = (percent) ' (base) [(product) = (factor) - (factor)]
+ + 4 4
80 z Q - 110
Q = 80+110 Divide.
Q = 0.727272... Round to two decimal places.
= 137 The symbol ”” is read as ” approximately.”
Thus, this bottle contains approximately 73% HCI.
Do [link], [link].
Example:
Five years ago a woman had an annual income of $19,200. She presently earns $42,000 annually. By what
percent has her salary increased? Round the result to two decimal places.
We need to determine the percent.
42,000 is what percent of 19,200? Missing factor statement.
il {=~ ——"— iL i
(percentage) = (percent) - (base)
4 4 i +
42,000 a Q - 19,200
Q = 42,000 + 19,200 Divide.
Q = Balls Round to two decimal places.
Q 2.19 Convert to a percent.
QO = 219% Convert to a percent.
Thus, this woman's annual salary has increased 219%.
Practice Set B
Exercise:
Problem: 99.13 is what percent of 431?
Solution:
23%
Exercise:
Problem:
On an 80 question exam, a student got 72 correct answers. What percent did the student get on the
exam?
Solution:
90%
Exercise:
Problem:
A bottle contains 45 milliliters of sugar and 67 milliliters of water. What fraction of sugar does the
bottle contain? Round the result to two decimal places (then express as a percent).
Solution:
40%
Finding the Base
Sample Set C
Example:
15 is 30% of what number? Missing factor statement.
i, 1 t NS capers
(percentage) = (percent) i (base) (percentage) = (factor) - (factor)|
+ t L u)
5 = 30% : B Convert to decimals.
15 = .30 : B (missing factor) = (product) + (known factor)|
B = 15+.30
B = 50
Thus, 15 is 30% of 50.
Try [link] in [link].
Example:
ue i ee e. what number? Missing factor statement.
eee
(percentage) = (percent) - (base)
4 4 4 + 4
56.43 = 33% . B Convert to decimals.
56.43 = 33 . B Divide.
B = 56.43 + .33
eel call
Thus, 56.43 is 33% of 171.
Try [link] in [link].
Example:
Fifteen milliliters of water represents 2% of a hydrochloric acid (HCl) solution. How many milliliters of
solution are there?
We need to determine the total supply. The word supply indicates base.
ue if 27 a what number? Missing factor statement.
SS
(percentage) = (percent) - (base)
4 4 4 + i
1155 = 2% : B Convert to decimals.
15 = .02 : B Divide.
B = 15+.02
B = 750
Thus, there are 750 milliliters of solution in the bottle.
Try Clink] in [link].
Example:
In a particular city, a sales tax of 65% is charged on items purchased in local stores. If the tax on an item is
$2.99, what is the price of the item?
We need to determine the price of the item. We can think of price as the starting place. Starting place
indicates base. We need to determine the base.
ae is 65% ce what number? Missing factor statement.
——
1
(percentage) = (percent) - (base)
i 4 i : f
2.99 = 65% B Convert to decimals.
2.99 = 6.5% B
2.99 = .065 183 [(missing factor) = (product) + (known factor)|
B = 2.99+ .065 Divide.
lf = 4G
Thus, the price of the item is $46.00.
Try [link] in [link].
Example:
A clothing item is priced at $20.40. This marked price includes a 15% discount. What is the original price?
We need to determine the original price. We can think of the original price as the starting place. Starting
place indicates base. We need to determine the base. The new price, $20.40, represents
100% — 15% = 85% of the original price.
20.40 i: 85% ae what number? Missing factor statement.
(percentage) = (percent) - (base)
4 4 4 + 4
20.40 = 85% : B Convert to decimals.
20.40 = .85 : B [(missing factor) = (product) + (known factor)|
B = 20.40 + .85 Divide.
(Bi — a4
Thus, the original price of the item is $24.00.
Try [link] in [link].
Practice Set C
Exercise:
Problem: 1.98 is 2% of what number?
Solution:
99
Exercise:
Problem:
3.3 milliliters of HCl represents 25% of an HCl solution. How many milliliters of solution are there?
Solution:
13.2ml
Exercise:
Problem:
A salesman, who makes a commission of 184% on each sale, makes a commission of $152.39 ona
particular sale. Rounded to the nearest dollar, what is the amount of the sale?
Solution:
$835
Exercise:
Problem:
At "super-long play," 2 5 hours of play of a video cassette recorder represents 31.25% of the total
playing time. What is the total playing time?
Solution:
8 hours
Exercises
For the following 25 problems, find each indicated quantity.
Exercise:
Problem: What is 21% of 104?
Solution:
21.84
Exercise:
Problem: What is 8% of 36?
Exercise:
Problem: What is 98% of 545?
Solution:
534.1
Exercise:
Problem: What is 143% of 33?
Exercise:
Problem: What is 104% of 20?
Solution:
2.1
Exercise:
Problem
Exercise:
Problem
: 3.25 is what percent of 88°?
: 22.44 is what percent of 44?
Solution:
51
Exercise:
Problem
Exercise:
Problem
: 0.0036 is what percent of 0.03?
: 31.2 is what percent of 26?
Solution:
120
Exercise:
Problem
Exercise:
Problem
: 266.4 is what percent of 74?
: 0.0101 is what percent of 0.0505?
Solution:
20
Exercise:
Problem
Exercise:
Problem
: 2.4 is 24% of what number?
: 24.19 is 41% of what number?
Solution:
59
Exercise:
Problem
: 61.12 is 16% of what number?
Exercise:
Problem:
Solution:
91
Exercise:
Problem:
Exercise:
Problem:
Solution:
9.15
Exercise:
Problem:
Exercise:
Problem:
Solution:
568
Exercise:
Problem:
Exercise:
Problem:
Solution:
1.19351
Exercise:
Problem:
Exercise:
Problem:
Solution:
250
82.81 is 91% of what number?
115.5 is 20% of what number?
43.92 is 480% of what number?
What is 85% of 62?
29.14 is what percent of 5.13?
0.6156 is what percent of 5.13?
What is 0.41% of 291.1?
26.136 is 121% of what number?
1,937.5 is what percent of 775?
Exercise:
Problem: 1 is what percent of 2,000?
Exercise:
Problem: 0 is what percent of 59?
Solution:
0
Exercise:
Problem:
An item of clothing is on sale for 10% off the marked price. If the marked price is $14.95, what is the
sale price? (Round to two decimal places.)
Exercise:
Problem:
A grocery clerk, who makes $365 per month, receives a 7% raise. How much is her new monthly
salary?
Solution:
390.55
Exercise:
Problem:
An item of clothing which originally sells for $55.00 is marked down to $46.75. What percent has it
been marked down?
Exercise:
Problem: On a 25 question exam, a student gets 21 correct. What percent is this?
Solution:
84
Exercise:
Problem: On a 45 question exam, a student gets 40%. How many questions did this student get correct?
Exercise:
Problem:
A vitamin tablet, which weighs 250 milligrams, contains 35 milligrams of vitamin C. What percent of
the weight of this tablet is vitamin C?
Solution:
14
Exercise:
Problem:
Five years ago a secretary made $11,200 annually. The secretary now makes $17,920 annually. By what
percent has this secretary's salary been increased?
Exercise:
Problem:
A baseball team wins 48 3% of all their games. If they won 78 games, how many games did they play?
Solution:
160
Exercise:
Problem:
A typist was able to increase his speed by 120% to 42 words per minute. What was his original typing
speed?
Exercise:
Problem:
A salesperson makes a commission of 12% on the total amount of each sale. If, in one month, she
makes a total of $8,520 in sales, how much has she made in commission?
Solution:
$1,022.40
Exercise:
Problem:
A salesperson receives a salary of $850 per month plus a commission of 85% of her sales. If, ina
particular month, she sells $22,800 worth of merchandise, what will be her monthly earnings?
Exercise:
Problem:
A man borrows $1150.00 from a loan company. If he makes 12 equal monthly payments of $130.60,
what percent of the loan is he paying in interest?
Solution:
36.28%
Exercise:
Problem:
The distance from the sun to the earth is approximately 93,000,000 miles. The distance from the sun to
Pluto is approximately 860.2% of the distance from the sun to the Earth. Approximately, how many
miles is Pluto from the sun?
Exercise:
Problem:
The number of people on food stamps in Maine in 1975 was 151,000. By 1980, the number had
decreased to 59,200. By what percent did the number of people on food stamps decrease? (Round the
result to the nearest percent.)
Solution:
61
Exercise:
Problem:
In Nebraska, in 1960, there were 734,000 motor-vehicle registrations. By 1979, the total had increased
by about 165.6%. About how many motor-vehicle registrations were there in Nebraska in 1979?
Exercise:
Problem:
From 1973 to 1979, in the United States, there was an increase of 166.6% of Ph.D. social scientists to
52,000. How many were there in 1973?
Solution:
19,500
Exercise:
Problem:
In 1950, in the United States, there were 1,894 daily newspapers. That number decreased to 1,747 by
1981. What percent did the number of daily newspapers decrease?
Exercise:
Problem:
A particular alloy is 27% copper. How many pounds of copper are there in 55 pounds of the alloy?
Solution:
14.85
Exercise:
Problem:
A bottle containing a solution of hydrochloric acid (HCl) is marked 15% (meaning that 15% of the HCl
solution is acid). If a bottle contains 65 milliliters of solution, how many milliliters of water does it
contain?
Exercise:
Problem:
A bottle containing a solution of HCl is marked 45%. A test shows that 36 of the 80 milliliters
contained in the bottle are hydrochloric acid. Is the bottle marked correctly? If not, how should it be
remarked?
Solution:
Marked correctly
Exercises For Review
Exercise:
Problem: ((link]) Use the numbers 4 and 7 to illustrate the commutative property of multiplication.
Exercise:
Problem: (|link|) Convert 4 to a mixed number.
Solution:
4
25
Exercise:
Problem: ((link]) Arrange the numbers +; > and 4 in increasing order.
Exercise:
Problem: ((link]) Convert 4.006 to a mixed number.
Solution:
3
4500
Exercise:
Problem: ((link]) Convert : % to a fraction.
Summary of Key Concepts
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module reviews the key concepts from the chapter
"Ratios and Rates."
Summary of Key Concepts
Denominate Numbers ([link])
Numbers that appear along with units are denominate numbers. The
amounts 6 dollars and 4 pints are examples of denominate numbers.
Like and Unlike Denominate Numbers ([link])
Like denominate numbers are denominate numbers with like units. If the
units are not the same, the numbers are unlike denominate numbers.
Pure Numbers ([(link])
Numbers appearing without a unit are pure numbers.
Comparing Numbers by Subtraction and Division (({link])
Comparison of two numbers by subtraction indicates how much more one
number is than another. Comparison by division indicates how many times
larger or smaller one number is than another.
Comparing Pure or Like Denominate Numbers by Subtraction ((link])
Numbers can be compared by subtraction if and only if they are pure
numbers or like denominate numbers.
Ratio Rate ({link])
A comparison, by division, of two like denominate numbers is a ratio. A
comparison, by division, of two unlike denominate numbers is a rate.
Proportion ([{link])
A proportion is a statement that two ratios or rates are equal.
3 people 6 people
Qjobs — ~djobs 'S 4 Proportion.
Solving a Proportion ([link])
To solve a proportion that contains three known numbers and a letter that
represents an unknown quantity, perform the cross multiplication, then
divide the product of the two numbers by the number that multiplies the
letter.
Proportions Involving Rates ((link])
When writing a proportion involving rates it is very important to write it so
that the same type of units appears on the same side of either the equal sign
or the fraction bar.
unit typel — unit type l unit typel — unit type2
unit type2 ~— unit type 2 unit typel ~~ unit type 2
Five-Step Method for Solving Proportions ([link])
1. By careful reading, determine what the unknown quantity is and
represent it with some letter. There will be only one unknown in a
problem.
2. Identify the three specified numbers.
3. Determine which comparisons are to be made and set up the
proportion.
4. Solve the proportion.
5. Interpret and write a conclusion.
When solving applied problems, ALWAYS begin by determining the
unknown quantity and representing it with a letter.
Percents ({link])
A ratio in which one number is compared to 100 is a percent. Percent
means "for each hundred."
Conversion of Fractions, Decimals, and Percents ({link])
It is possible to convert decimals to percents, fractions to percents, percents
to decimals, and percents to fractions.
Applications of Percents:
The three basic types of percent problems involve a base, a percentage,
and a percent.
Base ((link])
The base is the number used for comparison.
Percentage ({link])
The percentage is the number being compared to the base.
Percent ((link])
By its definition, percent means part of.
Solving Problems ({link])
Percentage = (percent) x (base)
Percent — percentage
base
Base = percentage
percent
Exercise Supplement
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is an exercise supplement for the chapter
"Ratios and Rates" and contains many exercise problems. Odd problems are
accompanied by solutions.
Exercise Supplement
Ratios and Rates ((link])
Exercise:
Problem:Compare 250 watts to 100 watts by subtraction.
Solution:
250 watts are 150 watts more than 100 watts
Exercise:
Problem: Compare 126 and 48 by subtraction.
Exercise:
Problem: Compare 98 radishes to 41 radishes by division.
Solution:
98 radishes are 2.39 times as many radishes as 41 radishes
Exercise:
Problem: Compare 144 to 9 by division.
Exercise:
Problem: Compare 100 tents to 5 tents by division.
Solution:
100 tents are 20 times as many tents as 5 tents
Exercise:
Problem: Compare 28 feet to 7 feet by division.
Exercise:
Problem:
Comparison, by division, of two pure numbers or two like denominate
numbers is calleda.
Solution:
ratio
Exercise:
Problem:
A comparison, by division, of two unlike denominate numbers is
calleda.
For problems 9-12, express each ratio or rate as a fraction.
Exercise:
Problem:15 to 5
Solution:
S.
1
Exercise:
Problem:72 to 12
Exercise:
Problem:8 millimeters to 5 milliliters
Solution:
8ml
5ml
Exercise:
Problem: 106 tablets to 52 tablets
For problems 13-16, write each ratio in the form "a to b".
Exercise:
Problem: Es
16
Solution:
9 to 16
Exercise:
3)
Problem: —
11
Exercise:
1 diskette
Problem: 8 diskettes
Solution:
1 diskette to 8 diskettes
Exercise:
5D papers
Problem: apes
For problems 17-21, write each ratio or rate using words.
Exercise:
9 18
Problem: — = —
roblem 16 39
Solution:
9 is to 16 as 18 is to 32
Exercise:
1 12
Problem: — = —
A A8
Exercise:
. S items _ 2 items
Problem: 7 aojars = 1 dollar
Solution:
8 items are to 4 dollars as 2 items are to 1 dollar
Exercise:
Problem:
150 milligrams of niacin is to 2 tablets as 300 milligrams of niacin is
to 4 tablets.
Exercise:
Problem: 20 people is to 4 seats as 5 people is to 1 seat.
Solution:
20 _ 5
Fae |
20 people are to 4 seats as 5 people are to 1 seat
Proportions ((link])
For problems 22-27, determine the missing number in each proportion.
Exercise:
24
Problem: ~ a
3
Exercise:
15 60
Problem: — = —
‘4 7
Solution:
28
Exercise:
Problem: — = =
44
Exercise:
3 15
Problem: — = —
50
Solution:
10
Exercise:
Problem: 15 bats =. bats
16 balls 128 balls
Exercise:
. 96 rooms __ 504 rooms
Problem: 29 fans —séi ‘: fans
Solution:
406
Applications of Proportions ({link])
Exercise:
Problem:
On a map, 3 inches represents 20 miles. How many miles does 27
inches represent?
Exercise:
Problem:
A salt solution is composed of 8 parts of salt to 5 parts of water. How
many parts of salt are there in a solution that contains 50 parts of
water?
Solution:
80
Exercise:
Problem:
4
A model is built to Tp scale. If a particular part of the model measures
8 inches in length, how long is the actual structure?
Exercise:
Problem:
The ratio of ammonia to air in a container is 40 How many milliliters
of air should be in a container that contains 8 milliliters of ammonia?
Solution:
30 or 1062
Exercise:
Problem:
A 4-foot girl casts a 9-foot shadow at a particular time of the day. How
tall is a pole that casts a 144-foot shadow at the same time of the day?
Exercise:
Problem:
The odds that a particular event will occur are 11 to 2. If this event
occurs 55 times, how many times would you predict it does not occur?
Solution:
10
Exercise:
Problem:
3
Every 1 e teaspoon of a multiple vitamin, in granular form, contains
0.85 the minimum daily requirement of vitamin A. How many
teaspoons of this vitamin are required to supply 2.25 the minimum
daily requirement?
Percent and Fractions of One Percent ((link],[link])
For problems 35-39, convert each decimal to a percent.
Exercise:
Problem:0.16
Solution:
16%
Exercise:
Problem:0.818
Exercise:
Problem:5.3536
Solution:
935.36%
Exercise:
Problem:0.50
Exercise:
Problem:3
Solution:
300%
For problems 40-48, convert each percent to a decimal.
Exercise:
Problem:62%
Exercise:
Problem: 1.58%
Solution:
0.0158
Exercise:
Problem:9.15%
Exercise:
Problem:0.06%
Solution:
0.0006
Exercise:
Problem:0.003%
Exercise:
3
Problem:5—— % to a three-place decimal
Solution:
0.053
Exercise:
9
Problem: 3 % to a three-place decimal
Exercise:
25
Problem:82 39 % to a four-place decimal
Solution:
0.8286
Exercise:
1
Problem: 18 7 % to a four-place decimal
For problems 49-55, convert each fraction or mixed number to a percent.
Exercise:
3
Problem: 5
Solution:
60%
Exercise:
Problem: —
ro em:
Exercise:
Problem: —
Solution:
31.25%
Exercise:
Problem: —
Exercise:
105
Problem: ——
Solution:
656.25%
Exercise:
1
Problem:45 —
roblem ii
Exercise:
278
Problem:6 or
Solution:
3688.8%
For problems 56-64, convert each percent to a fraction or mixed number.
Exercise:
Problem:95%
Exercise:
Problem: 12%
Solution:
3
25
Exercise:
Problem:83%
Exercise:
Problem:38.125%
Solution:
61
160
Exercise:
Problem:61. 2%
Exercise:
5
Problem: %
Solution:
ee
160
Exercise:
9
Problem: 6 —
roblem oa %
Exercise:
3
Problem: 15 — %
22
Solution:
2977
19800
Exercise:
19
Problem: 106 —
roblem iE %
Applications of Percents ({link])
For problems 65-72, find each solution.
Exercise:
Problem: What is 16% of 40?
Solution:
6.4
Exercise:
Problem: 29.4 is what percent of 105°?
Exercise:
Pall
POE ei is 547.2% of what number?
Solution:
0.625 or 2
Exercise:
Problem:0.09378 is what percent of 52.1?
Exercise:
Problem: What is 680% of 1.41?
Solution:
9.588
Exercise:
Problem:
A kitchen knife is on sale for 15% off the marked price. If the marked
price is $ 39.50, what is the sale price?
Exercise:
Problem:
On an 80 question geology exam, a student gets 68 correct. What
percent is correct?
Solution:
85
Exercise:
Problem:
A salesperson makes a commission of 18% of her monthly sales total.
She also receives a monthly salary of $1,600.00. If, in a particular
month, she sells $4,000.00 worth of merchandise, how much will she
make that month?
Proficiency Exam
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is a proficiency exam to the chapter "Ratios
and Rates." Each problem is accompanied with a reference link pointing
back to the module that discusses the type of problem demonstrated in the
question. The problems in this exam are accompanied by solutions.
Proficiency Exam
Exercise:
Problem: ({link]) Compare 4 cassette tapes to 7 dollars.
Solution:
4 cassette tapes
7 dollars
Exercise:
Problem:
({link]) What do we call a comparison, by division, of two unlike
denominate numbers?
Solution:
Rate
For problems 3 and 4, express each ratio or rate as a fraction.
Exercise:
Problem: ({link]) 11 to 9
Solution:
dd
9
Exercise:
Problem: ({link]) 5 televisions to 2 radios
Solution:
5 televisions
2 radios
For problems 5 and 6, write each ratio or rate in the form "a to 6."
Exercise:
Problem: ((link]) tae
Solution:
8 maps to 3 people
Exercise:
. 2 psychologists
Problem: ((link!) oe
Solution:
two psychologists to seventy-five people
For problems 7-9, solve each proportion.
Exercise:
Problem: ({link]) & = a
Solution:
15
Exercise:
Problem: (({link]) = = -
Solution:
1
Exercise:
. . 3computers __ 24 computers
Problem: ({link]) 8students ~ «zx students
Solution:
64
Exercise:
Problem:
({link]) On a map, 4 inches represents 50 miles. How many miles does
3 inches represent?
Solution:
1
ot =
Exercise:
Problem:
({link]) An acid solution is composed of 6 milliliters of acid to 10
milliliters of water. How many milliliters of acid are there in an acid
solution that is composed of 3 milliliters of water?
Solution:
1.8
Exercise:
Problem:
({link]) The odds that a particular event will occur are 9 to 7. If the
event occurs 27 times, how many times would you predict it will it not
occur?
Solution:
21
For problems 13 and 14, convert each decimal to a percent.
Exercise:
Problem: ({link]) 0.82
Solution:
82%
Exercise:
Problem: ({link]) 5.7
Solution:
377 2%
For problems 15 and 16, convert each percent to a decimal.
Exercise:
Problem: ({link]) 2.813%
Solution:
0.02813
Exercise:
Problem: ({link]) 0.006%
Solution:
0.00006
For problems 17-19, convert each fraction to a percent.
Exercise:
Problem: ({link]) 2
Solution:
840%
Exercise:
Problem: ((link]) +
Solution:
12.5%
Exercise:
syne 800
Problem: ((link]) <>
Solution:
1,000%
For problems 20 and 21, convert each percent to a fraction.
Exercise:
Problem: ({link]) 15%
Solution:
3
20
Exercise:
Problem: ({link]) 3+ %
Solution:
4 1
2,700 Of 675
For problems 22-25, find each indicated quantity.
Exercise:
Problem: ({link]) What is 18% of 26?
Solution:
4.68
Exercise:
Problem: ({link]) 0.618 is what percent of 0.3?
Solution:
206
Exercise:
Problem: ({link]) 0.1 is 1.1% of what number?
Solution:
9.09
Exercise:
Problem:
({link]) A salesperson makes a monthly salary of $1,000.00. He also
gets a commission of 12% of his total monthly sales. If, in a particular
month, he sells $5,500.00 worth of merchandise, what is his income
that month?
Solution:
$1,660
Objectives
This module contains the learning objectives for the chapter "Techniques of
Estimation" from Fundamentals of Mathematics by Denny Burzynski and
Wade Ellis, jr.
After completing this chapter, you should
Estimation by Rounding (({link])
e understand the reason for estimation
e be able to estimate the result of an addition, multiplication, subtraction,
or division using the rounding technique
Estimation by Clustering ({link])
e understand the concept of clustering
e be able to estimate the result of adding more than two numbers when
clustering occurs using the clustering technique
Mental Arithmetic—Using the Distributive Property ({link])
e understand the distributive property
¢ be able to obtain the exact result of a multiplication using the
distributive property
Estimation by Rounding Fractions ({link])
e be able to estimate the sum of two or more fractions using the
technique of rounding fractions
Estimation by Rounding
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to estimate by rounding. By
the end of the module students should understand the reason for estimation
and be able to estimate the result of an addition, multiplication, subtraction,
or division using the rounding technique.
Section Overview
e Estimation By Rounding
When beginning a computation, it is valuable to have an idea of what value
to expect for the result. When a computation is completed, it is valuable to
know if the result is reasonable.
In the rounding process, it is important to note two facts:
1. The rounding that is done in estimation does not always follow the
rules of rounding discussed in [link] (Rounding Whole Numbers).
Since estimation is concerned with the expected value of a
computation, rounding is done using convenience as the guide rather
than using hard-and-fast rounding rules. For example, if we wish to
estimate the result of the division 80 + 26, we might round 26 to 20
rather than to 30 since 80 is more conveniently divided by 20 than by
30.
2. Since rounding may occur out of convenience, and different people
have different ideas of what may be convenient, results of an
estimation done by rounding may vary. For a particular computation,
different people may get different estimated results. Results may vary.
Estimation
Estimation is the process of determining an expected value of a
computation.
Common words used in estimation are about, near, and between.
Estimation by Rounding
The rounding technique estimates the result of a computation by rounding
the numbers involved in the computation to one or two nonzero digits.
Sample Set A
Example:
Estimate the sum: 2,357 + 6,106.
Notice that 2,357 is near 2,400, and that 6,106 is near 6,100.
two nonzero two nonzero
digits digits
The sum can be estimated by 2,400 + 6,100 = 8,500. (It is quick and easy
to add 24 and 61.)
Thus, 2,357 + 6,106 is about 8,400. In fact, 2,357 + 6,106 = 8,463.
Practice Set A
Exercise:
Problem: Estimate the sum: 4,216 + 3,942.
Solution:
4,216 + 3,942 : 4,200 + 3,900. About 8,100. In fact, 8,158.
Exercise:
Problem: Estimate the sum: 812 + 514.
Solution:
812+ 514 : 800 + 500. About 1,300. In fact, 1,326.
Exercise:
Problem: Estimate the sum: 43,892 + 92,106.
Solution:
43,892 + 92,106 : 44,000 + 92,000. About 136,000. In fact, 135,998.
Sample Set B
Example:
Estimate the difference: 5,203 — 3,015.
Notice that 5,203 is near 5,200, andthat 3,015 is near 3,000.
two nonzero one nonzero
digits digit
The difference can be estimated by 5,200 — 3,000 = 2,200.
Thus, 5,203 — 3,015 is about 2,200. In fact, 5,203 — 3,015 = 2,188.
We could make a less accurate estimation by observing that 5,203 is near
5,000. The number 5,000 has only one nonzero digit rather than two (as
does 5,200). This fact makes the estimation quicker (but a little less
accurate). We then estimate the difference by 5,000 — 3,000 = 2,000, and
conclude that 5,203 — 3,015 is about 2,000. This is why we say "answers
may vary."
Practice Set B
Exercise:
Problem: Estimate the difference: 628 — 413.
Solution:
628 — 413 : 600 — 400. About 200. In fact, 215.
Exercise:
Problem: Estimate the difference: 7,842 — 5,209.
Solution:
7,842 — 5,209 : 7,800 — 5,200. About 2,600. In fact, 2,633.
Exercise:
Problem: Estimate the difference: 73,812 — 28,492.
Solution:
73,812 — 28,492 : 74,000 — 28,000. About 46,000. In fact, 45,320.
Sample Set C
Example:
Estimate the product: 73 - 46.
Notice that 73isnear 70, andthat46isnear 50.
one nonzero one nonzero
digit digit
The product can be estimated by 70 - 50 = 3,500. (Recall that to multiply
numbers ending in zeros, we multiply the nonzero digits and affix to this
product the total number of ending zeros in the factors. See [link] for a
review of this technique.)
Thus, 73 - 46 is about 3,500. In fact, 73 - 46 = 3,358.
Example:
Estimate the product: 87 - 4,316.
Notice that 87 is closeto 90, andthat4,316iscloseto 4,000.
one tae one digit ero
The product can be estimated by 90 - 4,000 = 360,000.
Thus, 87 - 4,316 is about 360,000. In fact, 87 - 4,316 = 375,492.
Practice Set C
Exercise:
Problem: Estimate the product: 31 - 87.
Solution:
31-87: 30-90. About 2,700. In fact, 2,697.
Exercise:
Problem: Estimate the product: 18 - 42.
Solution:
18 - 42 : 20 - 40. About 800. In fact, 756.
Exercise:
Problem: Estimate the product: 16 - 94.
Solution:
16-94: 15-100. About 1,500. In fact, 1,504.
Sample Set D
Example:
Estimate the quotient: 153 + 17.
Notice that 153 iscloseto 150, andthat17iscloseto 165.
two nonzero
two nonzero ave
digits
digits
The quotient can be estimated by 150 + 15 = 10.
Thus, 153 + 17 is about 10. In fact, 153 + 17 = 9.
Example:
Estimate the quotient: 742,000 + 2,400.
Notice that 742,000 is close to 700,000 , and that 2,400 is close to
one nonzero
digit
2,000.
one nonzero
digit
The quotient can be estimated by 700,000 + 2,000 = 350.
Thus, 742,000 + 2,400 is about 350. In fact, 742,000 + 2,400 = 309.16.
Practice Set D
Exercise:
Problem: Estimate the quotient: 221 + 18.
Solution:
221 + 18: 200 = 20. About 10. In fact, 12.27.
Exercise:
Problem: Estimate the quotient: 4,079 + 381.
Solution:
4,079 + 381 : 4,000 + 400. About 10. In fact, 10.70603675...
Exercise:
Problem: Estimate the quotient: 609,000 + 16,000.
Solution:
609,000 + 16,000 : 600,000 + 15,000. About 40. In fact, 38.0625.
Sample Set E
Example:
Estimate the sum: 53.82 + 41.6.
Notice that 53.82 is closeto 54, andthat41.6iscloseto 42.
two nonzero two nonzero
digits digits
The sum can be estimated by 54 + 42 = 96.
Thus, 53.82 + 41.6 is about 96. In fact, 53.82 + 41.6 = 95.42.
Practice Set E
Exercise:
Problem: Estimate the sum: 61.02 + 26.8.
Solution:
61.02 + 26.8 : 61 + 27. About 88. In fact, 87.82.
Exercise:
Problem: Estimate the sum: 109.12 + 137.88.
Solution:
109.12 + 137.88 : 110 + 138. About 248. In fact, 247. We could have
estimated 137.88 with 140. Then 110 + 140 is an easy mental
addition. We would conclude then that 109.12 + 137.88 is about 250.
Sample Set F
Example:
Estimate the product: (31.28)(14.2).
Notice that 31.28 is closeto 30, andthat14.2iscloseto 10.
one nonzero two nonzero
digit digits
The product can be estimated by 30-15 = 450. (3-15 = 45, then affix
one zero.)
Thus, (31.28)(14.2) is about 450. In fact, (31.28)(14.2) = 444.176.
Example:
Estimate 21% of 5.42.
Notice that 21% = .21 as a decimal, andthat .21iscloseto 2.
one nonzero
digit
Notice also that 5.42 is close to 5:
one nonzero
digit
Then, 21% of 5.42 can be estimated by (.2)(5) = 1.
Thus, 21% of 5.42 is about 1. In fact, 21% of 5.42 is 1.1382.
Practice Set F
Exercise:
Problem: Estimate the product: (47.8)(21.1).
Solution:
(47.8) (21.1) : (50) (20). About 1,000. In fact, 1,008.58.
Exercise:
Problem: Estimate 32% of 14.88.
Solution:
32% of 14.88 : (.3)(15). About 4.5. In fact, 4.7616.
Exercises
Estimate each calculation using the method of rounding. After you have
made an estimate, find the exact value and compare this to the estimated
result to see if your estimated value is reasonable. Results may vary.
Exercise:
Problem: 1,402 + 2,198
Solution:
about 3,600; in fact 3,600
Exercise:
Problem: 3,481 + 4,216
Exercise:
Problem: 921 + 796
Solution:
about 1,700; in fact 1,717
Exercise:
Problem: 611 + 806
Exercise:
Problem: 4,681 + 9,325
Solution:
about 14,000; in fact 14,006
Exercise:
Problem: 6,476 + 7,814
Exercise:
Problem: 7,805 — 4,266
Solution:
about 3,500; in fact 3,539
Exercise:
Problem: 8,427 — 5,342
Exercise:
Problem: 14,106 — 8,412
Solution:
about 5,700; in fact 5,694
Exercise:
Problem: 26,486 — 18,931
Exercise:
Problem: 32 - 53
Solution:
about 1,500; in fact 1,696
Exercise:
Problem: 67 - 42
Exercise:
Problem: 628 - 891
Solution:
about 540,000; in fact 559,548
Exercise:
Problem: 426 - 741
Exercise:
Problem: 18,012 - 32,416
Solution:
about 583,200,000; in fact 583,876,992
Exercise:
Problem: 22,481 - 51,076
Exercise:
Problem: 287—19
Solution:
about 15; in fact 15.11
Exercise:
Problem: 884—33
Exercise:
Problem: 1,254—57
Solution:
about 20; in fact 22
Exercise:
Problem: 2,189~42
Exercise:
Problem: 8,092-239
Solution:
about 33; in fact 33.86
Exercise:
Problem: 2,688—48
Exercise:
Problem: 72.14 + 21.08
Solution:
about 93.2; in fact 93.22
Exercise:
Problem: 43.016 + 47.58
Exercise:
Problem: 96.53 — 26.91
Solution:
about 70; in fact 69.62
Exercise:
Problem: 115.0012 — 25.018
Exercise:
Problem: 206.19 + 142.38
Solution:
about 348.6; in fact 348.57
Exercise:
Problem: 592.131 + 211.6
Exercise:
Problem: (32.12) (48.7)
Solution:
about 1,568.0; in fact 1,564.244
Exercise:
Problem: (87.013)(21.07)
Exercise:
Problem: (3.003) (16.52)
Solution:
about 49.5; in fact 49.60956
Exercise:
Problem: (6.032)(14.091)
Exercise:
Problem: (114.06) (384.3)
Solution:
about 43,776; in fact 43,833.258
Exercise:
Problem: (5,137.118) (263.56)
Exercise:
Problem: (6.92)(0.88)
Solution:
about 6.21; in fact 6.0896
Exercise:
Problem: (83.04) (1.03)
Exercise:
Problem: (17.31)(.003)
Solution:
about 0.0519; in fact 0.05193
Exercise:
Problem: (14.016) (.016)
Exercise:
Problem: 93% of 7.01
Solution:
about 6.3; in fact 6.5193
Exercise:
Problem: 107% of 12.6
Exercise:
Problem: 32% of 15.3
Solution:
about 4.5; in fact 4.896
Exercise:
Problem: 74% of 21.93
Exercise:
Problem: 18% of 4.118
Solution:
about 0.8; in fact 0.74124
Exercise:
Problem: 4% of .863
Exercise:
Problem: 2% of .0039
Solution:
about 0.00008; in fact 0.000078
Exercises for Review
Exercise:
Problem: ({link]) Find the difference: = — 2.
Exercise:
1
6=—
ay .
Problem: ({link]) Find the value oa
4
Solution:
23
25
Exercise:
Problem: ({link]) Convert the complex decimal 1.11 to a decimal.
Exercise:
Problem:
({link]) A woman 5 foot tall casts an 8-foot shadow at a particular time
of the day. How tall is a tree that casts a 96-foot shadow at the same
time of the day?
Solution:
60 feet tall
Exercise:
Problem: ({link]) 11.62 is 83% of what number?
Estimation by Clustering
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to estimate by clustering. By
the end of the module students should understand the concept of clustering
and be able to estimate the result of adding more than two numbers when
clustering occurs using the clustering technique.
Section Overview
e Estimation by Clustering
Cluster
When more than two numbers are to be added, the sum may be estimated
using the clustering technique. The rounding technique could also be used,
but if several of the numbers are seen to cluster (are seen to be close to)
one particular number, the clustering technique provides a quicker estimate.
Consider a sum such as
32 + 68 + 29 + 73
Notice two things:
1. There are more than two numbers to be added.
2. Clustering occurs.
a. Both 68 and 73 cluster around 70, so 68 + 73 is close to
80 + 70 = 2(70) = 140.
ee eo
32 + 68 + 29+ 71
nee |
b. Both 32 and 29 cluster around 30, so 32 + 29 is close to
30 + 30 = 2(30) = 60.
The sum may be estimated by
(2-30) +(2-70) = 6+140
200
In fact, 32 + 68 + 29 + 73 = 202.
Sample Set A
Estimate each sum. Results may vary.
Example:
27 + 48+ 31+ 52.
27 and 31 cluster near 30. Their sum is about 2 - 30 = 60.
48 and 52 cluster near 50. Their sum is about 2 - 50 = 100.
2-30 2-50) = 60+ 100
Thus, 27 + 48 +31 + 52is about ‘ yaa ) 7 et
In fact, 27 + 48 + 31 + 52 = 158.
Example:
88 + 214+ 19+ 91.
88 and 91 cluster near 90. Their sum is about 2 - 90 = 180.
21 and 19 cluster near 20. Their sum is about 2 - 20 = 40.
2-90 2-20) = 180+ 40
Thus, 88421: 19 +91 is about‘ aus ) = ae
finetact: 6 Seo 2 er Ole 21:
Example:
17+ 21+ 48 + 18.
17, 21, and 18 cluster near 20. Their sum is about 3 - 20 = 60.
A8 is about 50.
3-20)+50 = 60+50
Thus, 17 + 21 + 48 + 18 is about ( ye ee
In fact, 17> 21-- 48 + 18: 104,
Example:
61+ 48+ 49+ 57+ 52.
61 and 57 cluster near 60. Their sum is about 2 - 60 = 120.
48, 49, and 52 cluster near 50. Their sum is about 3 - 50 = 150.
Thus, 61 + 48 + 49 + 57 + 52 is about
(2-60) + (3-50) = 1204150
=a
In fact, 61 + 48 + 49 + 57 + 52 = 267.
Example:
706 + 321 + 293 + 684.
706 and 684 cluster near 700. Their sum is about 2 - 700 = 1,400.
321 and 293 cluster near 300. Their sum is about 2 - 300 = 600.
Thus, 706 + 321 + 293 + 684 is about
(2-700) + (2-300) = 1,400 + 600
=e 00
In fact, 706 + 321 + 293 + 684 = 2,004.
Practice Set A
Use the clustering method to estimate each sum.
Exercise:
Problem: 28 + 51+ 31+ 47
Solution:
(2 - 30) + (2-50) — 60 + 100 = 160
Exercise:
Problem: 42 + 39 + 68 + 41
Solution:
(3-40) + 70 = 120+ 70 = 190
Exercise:
Problem: 37 + 39 + 83 + 42+ 79
Solution:
(3 . 40) -- (2 . 80) — 120+ 160 = 280
Exercise:
Problem: 612 + 585 + 830 + 794
Solution:
(2 - 600) + (2 - 800) = 1,200 + 1,600 = 2,800
Exercises
Use the clustering method to estimate each sum. Results may vary.
Exercise:
Problem: 28 + 51+ 31+ 47
Solution:
2(30) + 2(50) = 160 (157)
Exercise:
Problem: 42 + 19 + 39 + 23
Exercise:
Problem: 88 + 62 + 59 + 90
Solution:
2(90) + 2(60) = 300 (299)
Exercise:
Problem: 76 + 29 + 33 + 82
Exercise:
Problem: 19 + 23 + 87+ 21
Solution:
3(20) + 90 = 150 (150)
Exercise:
Problem: 41 + 28 + 42 + 37
Exercise:
Problem: 89 + 32 + 89 + 93
Solution:
3(90) + 30 = 300 (303)
Exercise:
Problem: 73 + 72 + 27+ 71
Exercise:
Problem: 43 + 62 + 61+ 55
Solution:
40 + 3(60) = 220 (221)
Exercise:
Problem: 31 + 77 + 31+ 27
Exercise:
Problem: 57 + 34+ 28+ 61+ 62
Solution:
3(60) + 2(30) = 240 (242)
Exercise:
Problem: 94 + 18 + 23+ 91+ 19
Exercise:
Problem: 103 + 72 + 66 + 97 + 99
Solution:
3(100) + 2(70) = 440 (437)
Exercise:
Problem: 42 + 1214+. 119+ 124+ 41
Exercise:
Problem: 19 + 24 + 87 + 23 + 91+ 93
Solution:
3(20) + 3(90) = 330 (337)
Exercise:
Problem: 108 + 61 + 63+ 96 + 57+ 99
Exercise:
Problem: 518 + 721 + 493 + 689
Solution:
2(500) + 2(700) = 2,400 (2,421)
Exercise:
Problem: 981 + 1208 + 1214 + 1006
Exercise:
Problem: 23 + 81+. 77+ 79+ 19+ 81
Solution:
2(20) + 4(80) = 360 (360)
Exercise:
Problem: 94 + 68 + 66 + 101+ 106 + 71+ 110
Exercises for Review
Exercise:
Problem: ([link]) Specify all the digits greater than 6.
Solution:
7, 8,9
Exercise:
Problem: ([link]) Find the product: + : = : i.
Exercise:
Problem: ({link]) Convert 0.06 to a fraction.
Solution:
3)
50
Exercise:
Problem:
({link]) Write the proportion in fractional form: "5 is to 8 as 25 is to
40."
Exercise:
Problem:
({link]) Estimate the sum using the method of rounding:
4,882 + 2,704.
Solution:
4,900 + 2,700 = 7,600 (7,586)
Mental Arithmetic-Using the Distributive Property
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This
module discusses using the distributive property. By the end of the module students should
understand the distributive property and be able to obtain the exact result of a multiplication using
the distributive property.
Section Overview
e The Distributive Property
e Estimation Using the Distributive Property
The Distributive Property
Distributive Property
The distributive property is a characteristic of numbers that involves both addition and
multiplication. It is used often in algebra, and we can use it now to obtain exact results for a
multiplication.
Suppose we wish to compute 3(2 + 5). We can proceed in either of two ways, one way which is
known to us already (the order of operations), and a new way (the distributive property).
1. Compute 3(2 + 5) using the order of operations.
3(2 + 5)
Operate inside the parentheses first: 2+ 5 = 7.
3(2+ 5) =3-7
Now multiply 3 and 7.
31245) =3-7=21
Thus, 3(2 + 5) = 21.
2. Compute 3(2 + 5) using the distributive property.
We know that multiplication describes repeated addition. Thus,
3(2+5) = 24542454245
2+ 5 appears 3 times
= 24242454545 (by the commutative property of addition)
= $-24+3-5 (since multiplication describes repeated addition)
6+ 15
21
Thus, 3(2 + 5) = 21.
Let's look again at this use of the distributive property.
3(2 + 5)
2-0 +2459 +245
2+ 5 appears 3 times
32245) = 286942 4 54545
2 appears 3 times 5 appears 3 times
= 3-2 + 3°5
Te ee RR ORS
3times2 3 times 5
The 3 has been distributed to the 2 and 5.
This is the distributive property. We distribute the factor to each addend in the parentheses.
The distributive property works for both sums and differences.
Sample Set A
Using the order of operations, we get
4(6+2) = 4-8
= 32
Using the order of operations, we get
8(9+6) = 8-15
= ulead
Example:
4(9—5) =4-9-4-5
Ce = 36 — 20
=16
Example:
25(20 — 3) = 25 - 20-25-38
= 500 — 75
= 425
Practice Set A
Use the distributive property to compute each value.
Exercise:
Problem: 6(8 + 4)
Solution:
6-8+6-4= 48+ 24= 72
Exercise:
Problem: 4(4 + 7)
Solution:
4-4+4-7=16+ 28= 44
Exercise:
Problem: 8(2 + 9)
Solution:
8-2+8-9=16+ 72 = 88
Exercise:
Problem: 12(10 + 3)
Solution:
12-10+12-3 = 120+ 36 = 156
Exercise:
Problem: 6(11 — 3)
Solution:
6-11—6-3 = 66 — 18 = 48
Exercise:
Problem: 8(9 — 7)
Solution:
8-9-8-7 = 72-56 = 16
Exercise:
Problem: 15(30 — 8)
Solution:
15-30 —15-8 = 450 — 120 = 330
Estimation Using the Distributive Property
We can use the distributive property to obtain exact results for products such as 25 - 23. The
distributive property works best for products when one of the factors ends in 0 or 5. We shall restrict
our attention to only such products.
Sample Set B
Use the distributive property to compute each value.
Example:
25 - 23
Notice that 23 = 20 + 3. We now write
25 * 23 = 25(20 +,3)
a
=25-20+25-3
= 500 + 75
= 575
hus, 25-23 — 575
We could have proceeded by writing 23 as 30 — 7.
25 - 23 = 25(30 — 7)
—
= 25+ 30—25-7
= 750 — 175
= 575
Example:
15 - 37
Notice that 37 = 30 + 7. We now write
15 - 37 = 15(30 +7)
se
=15-304+15-7
= 450 + 105
= 555
Thus, 15 - 37 = 555
We could have proceeded by writing 37 as 40 — 3.
15 - 37 = 15(40 — 3)
aa
=15-40-15-3
= 600 — 45
= 555
Example:
15 - 86
Notice that 86 = 80 + 6. We now write
15 - 86 = 15(80 +6)
oo
=15-80+15°6
= 1,200 + 90
= 1,290
We could have proceeded by writing 86 as 90 — 4.
15 - 86 = 15(90 — 4)
<a
=15-90—15-4
= 1,350 — 60
= 1,290
Practice Set B
Use the distributive property to compute each value.
Exercise:
Problem: 25 - 12
Solution:
25(10 + 2) = 25-10 + 25-2 = 250 + 50 = 300
Exercise:
Problem: 35 - 14
Solution:
35(10 + 4) = 35-10+ 35-4 = 350 + 140 = 490
Exercise:
Problem: 80 - 58
Solution:
80(50 + 8) = 80-50 + 80-8 = 4,000 + 640 = 4,640
Exercise:
Problem: 65 - 62
Solution:
65(60 + 2) = 65-60 + 65 - 2 = 3,900 + 130 = 4,030
Exercises
Use the distributive property to compute each product.
Exercise:
Problem: 15 - 13
Solution:
15(10 + 3) = 150 + 45 = 195
Exercise:
Problem: 15 - 14
Exercise:
Problem: 25 - 11
Solution:
25(10 + 1) = 250 + 25 = 275
Exercise:
Problem: 25 - 16
Exercise:
Problem: 15 - 16
Solution:
15(20 — 4) = 300 — 60 = 240
Exercise:
Problem: 35 - 12
Exercise:
Problem: 45 - 83
Solution:
45(80 + 3) = 3600 + 135 = 3735
Exercise:
Problem: 45 - 38
Exercise:
Problem: 25 - 38
Solution:
25(40 — 2) = 1,000 — 50 = 950
Exercise:
Problem: 25 - 96
Exercise:
Problem: 75 - 14
Solution:
75(10 + 4) = 750 + 300 = 1,050
Exercise:
Problem: 85 - 34
Exercise:
Problem: 65 - 26
Solution:
65(20 + 6) = 1,300 + 390 = 1,690 or 65(30 — 4) = 1,950 — 260 = 1,690
Exercise:
Problem: 55 - 51
Exercise:
Problem: 15 - 107
Solution:
15(100 + 7) = 1,500 + 105 = 1,605
Exercise:
Problem: 25 - 208
Exercise:
Problem: 35 - 402
Solution:
35(400 + 2) = 14,000 + 70 = 14,070
Exercise:
Problem: 85 - 110
Exercise:
Problem: 95 - 12
Solution:
95(10 + 2) = 950 + 190 = 1,140
Exercise:
Problem: 65 - 40
Exercise:
Problem: 80 - 32
Solution:
80(30 + 2) = 2,400 + 160 = 2,560
Exercise:
Problem: 30 - 47
Exercise:
Problem: 50 - 63
Solution:
50(60 + 3) = 3,000 + 150 = 3,150
Exercise:
Problem: 90 - 78
Exercise:
Problem: 40 - 89
Solution:
40(90 — 1) = 3,600 — 40 = 3,560
Exercises for Review
Exercise:
Problem: (([link]) Find the greatest common factor of 360 and 3,780.
Exercise:
Problem: ((link]) Reduce eee to lowest terms.
Solution:
x
26
Exercise:
Problem: ((link]) 13 of 27 is what number?
Exercise:
Problem: ((link]) Solve the proportion: dk =o:
Solution:
= 42
Exercise:
Problem: ((link]) Use the clustering method to estimate the sum: 88 + 106 + 91 + 114.
Estimation by Rounding Fractions
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses
how to estimate by rounding fractions. By the end of the module students should be able to estimate the sum of
two or more fractions using the technique of rounding fractions.
Section Overview
e Estimation by Rounding Fractions
Estimation by rounding fractions is a useful technique for estimating the result of a computation involving
fractions. Fractions are commonly rounded to + + 3. 0, and 1. Remember that rounding may cause estimates to
vary.
Sample Set A
Make each estimate remembering that results may vary.
Example:
‘ 3 5
Estimate 5 oP Spy
Notice that 3 is about +, and that + is about +.
3 ie ie eles 3 5 Gl ai,
Thus, 5 + 7g is about z+ = 1. In fact, ear ip = wee little more than 1.
Example:
Estimate 52 AF 45 ar 11z.
Adding the whole number parts, we get 20. Notice that = is close to = “S is close to 1, and = is close to =
TMS ee ech arc alee eee ila s 15.
Thus, 53 + 42 + 11+ is close to 20+ 15 = 215.
In fact, 52 + 45% + 112 = 21-2, alittle less than 215.
Practice Set A
Use the method of rounding fractions to estimate the result of each computation. Results may vary.
Exercise:
ib 5
Problem: gt ty
Solution:
1 1 2 1
Results may vary. + + + = 1. In fact, 2 + + = = =l1z5;
Exercise:
re eae
Problem: 91s
Solution:
Results may vary. 1 + > = ie In fact, t += iz
Exercise:
.Q4 7
Problem: rey a 355
Solution:
Results may vary. 84 +33 =11+1=12. In fact, 84 +30 =112
Exercise:
-16L it
Problem: 1655 + 43
Solution:
Results may vary. (16 + 0) + (4+ 1) = 16+5 = 21. In fact, 1655 4 4t = 20 ar
Exercises
Estimate each sum or difference using the method of rounding. After you have made an estimate, find the exact
value of the sum or difference and compare this result to the estimated value. Result may vary.
Exercise:
5 7
Problem: eo ug:
Solution:
1+1=2 (132)
Exercise:
ao
Problem: g tig
Exercise:
. 9 3
Problem: 19
Solution:
1 1/41
1+ 34 =13(19)
Exercise:
Problem: i= eek
Exercise:
Problem: _ eee
Solution:
Exercise:
sa 4
Problem: ig t 5
Exercise:
Problem: ca oe te
Solution:
14+0=1(14)
Exercise:
- 29 4 Ad
Problem: 30 + 0
Exercise:
Problem: -> + 6-4
Solution:
1 1 103
z +63 =7 (633)
Exercise:
Problem: 2 sg
15
Exercise:
re 3
Problem: i + 2 3
Solution:
1+25 =35(3%)
Exercise:
. 19 5
Problem: at 1G
Exercise:
293 1
Problem: 8 5 ot 4 5
Solution:
85 +4 = 125 (1233)
Exercise:
Problem: 53 + 25
Exercise:
Problem: 975 + 6+
Solution:
9+7=16 (1533)
Exercise:
.75 1
Problem: 755 + 10
Exercise:
Problem: 35 + 25 “= 1z
Solution:
35 +25+2=8 (730)
Exercise:
Problem: 6+ + 1a + 52
Exercise:
. 15 7
Problem: nos
Solution:
1
1—1=0 (=)
Exercise:
Problem: = — aa
Exercises for Review
Exercise:
Problem:
({link]) The fact that
(a first number - a second number) - a third number = a first number - (a second number - a third num
is an example of which property of multiplication?
Solution:
associative
Exercise:
Problem: ({link]) Find the quotient: = > =.
Exercise:
Problem: ((link]) Find the difference: 32 — 24.
Solution:
&
9
Exercise:
Problem: ({link]) Find the quotient: 4.6 + 0.11.
Exercise:
Problem: ((link]) Use the distributive property to compute the product: 25 - 37.
Solution:
25(40 — 3) = 1000 — 75 = 925
Summary of Key Concepts
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module reviews the key concepts from the chapter
"Techniques of Estimation."
Summary of Key Concepts
Estimation ((link])
Estimation is the process of determining an expected value of a
computation.
Estimation By Rounding ([link])
The rounding technique estimates the result of a computation by rounding
the numbers involved in the computation to one or two nonzero digits. For
example, 512 + 896 can be estimated by 500 + 900 = 1,400.
Cluster ({link])
When several numbers are close to one particular number, they are said to
cluster near that particular number.
Estimation By Clustering ({link])
The clustering technique of estimation can be used when
1. there are more than two numbers to be added, and
2. clustering occurs.
For example, 31 + 62 + 28 + 59 can be estimated by
(2 - 30) + (2-60) = 60 + 120 = 180
Distributive Property ([link])
The distributive property is a characteristic of numbers that involves both
addition and multiplication. For example,
3(4+ 6) =3-44+3-6=12+4 18 = 30
Estimation Using the Distributive Property ((link])
The distributive property can be used to obtain exact results for a
multiplication.
For example,
15-23 = 15- (20+ 3) = 15-204 15-3 = 300-4 45 = 345
Estimation by Rounding Fractions ([link])
Estimation by rounding fractions commonly rounds fractions to . = 4,
0, and 1.
For example,
5 5 1 =
a5. + Gg cam be estimated by = + 4 = >
Exercise Supplement
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is an exercise supplement for the chapter
"Techniques of Estimation" and contains many exercise problems. Odd
problems are accompanied by solutions.
Exercise Supplement
Estimation by Rounding ((link])
For problems 1-70, estimate each value using the method of rounding. After
you have made an estimate, find the exact value. Compare the exact and
estimated values. Results may vary.
Exercise:
Problem: 286 + 312
Solution:
600 (598)
Exercise:
Problem: 419 + 582
Exercise:
Problem: 689 + 511
Solution:
(1,200)
Exercise:
Problem: 926 + 1,105
Exercise:
Problem: 1,927 + 3,017
Solution:
4,900 (4,944)
Exercise:
Problem: 5,026 + 2,814
Exercise:
Problem: 1,408 + 2,352
Solution:
3,800 (3,760)
Exercise:
Problem: 1,186 + 4,228
Exercise:
Problem: 5,771 + 246
Solution:
6,050 (6,017)
Exercise:
Problem: 8,305 + 484
Exercise:
Problem: 3,812 + 2,906
Solution:
6,700 (6,718)
Exercise:
Problem: 5,293 + 8,007
Exercise:
Problem: 28,481 + 32,856
Solution:
61,400 (61,337)
Exercise:
Problem: 92,512 + 26,071
Exercise:
Problem: 87,612 + 2,106
Solution:
89,700 (89,718)
Exercise:
Problem: 42,612 + 4,861
Exercise:
Problem: 212,413 + 609
Solution:
213,000 (213,022)
Exercise:
Problem: 487,235 + 494
Exercise:
Problem: 2,409 + 1,526
Solution:
3,900 (3,935)
Exercise:
Problem: 3,704 + 4,704
Exercise:
Problem: 41 - 63
Solution:
2,400 (2,583)
Exercise:
Problem: 38 - 81
Exercise:
Problem: 18 - 28
Solution:
600 (504)
Exercise:
Problem: 52 - 21
Exercise:
Problem: 307 - 489
Solution:
150,123 147,000 (150,123)
Exercise:
Problem: 412 - 807
Exercise:
Problem: 77 - 614
Solution:
47,278 48,000 (47,278)
Exercise:
Problem: 62 - 596
Exercise:
Problem: 27 - 473
Solution:
12,771 14,100 (12,711)
Exercise:
Problem: 92 - 336
Exercise:
Problem: 12 - 814
Solution:
8,100 (9,768)
Exercise:
Problem: 8 - 2,106
Exercise:
Problem: 192 - 452
Solution:
90,000 (86,784)
Exercise:
Problem: 374 - 816
Exercise:
Problem: 88 - 4,392
Solution:
396,000 (386,496)
Exercise:
Problem: 126 - 2,834
Exercise:
Problem: 3,896 - 413
Solution:
1,609,048 1,560,000 (1,609,048)
Exercise:
Problem: 5,794 - 837
Exercise:
Problem: 6,311 - 3,512
Solution:
22,050,000 (22,164,232)
Exercise:
Problem: 7,471 - 5,782
Exercise:
Problem: 180 = 12
Solution:
18 (15)
Exercise:
Problem: 309 — 16
Exercise:
Problem: 286 — 22
Solution:
144 (13)
Exercise:
Problem: 527 — 17
Exercise:
Problem: 1,007 + 19
Solution:
50 (53)
Exercise:
Problem: 1,728 ~ 36
Exercise:
Problem: 2,703 ~ 53
Solution:
54 (51)
Exercise:
Problem: 2,562 + 61
Exercise:
Problem: 1,260 + 12
Solution:
130 (105)
Exercise:
Problem: 3,618 + 18
Exercise:
Problem: 3,344 + 76
Solution:
41.25 (44)
Exercise:
Problem: 7,476 ~ 356
Exercise:
Problem: 20,984 ~ 488
Solution:
42 (43)
Exercise:
Problem: 43,776 ~ 608
Exercise:
Problem: 7,196 ~ 514
Solution:
14.4 (14)
Exercise:
Problem: 51,492 + 514
Exercise:
Problem: 26,962 + 442
Solution:
60 (61)
Exercise:
Problem: 33,712 + 112
Exercise:
Problem: 105,152 ~ 106
Solution:
1,000 (992)
Exercise:
Problem: 176,978 ~ 214
Exercise:
Problem: 48.06 + 23.11
Solution:
71.1 (71.17)
Exercise:
Problem: 73.73 + 72.9
Exercise:
Problem: 62.91 + 56.4
Solution:
119.4 (119.31)
Exercise:
Problem: 87.865 + 46.772
Exercise:
Problem: 174.6 + 97.2
Solution:
272 (271.8)
Exercise:
Problem: (48.3) (29.6)
Exercise:
Problem: (87.11) (23.2)
Solution:
2,001 (2,020.952)
Exercise:
Problem: (107.02) (48.7)
Exercise:
Problem: (0.76) (5.21)
Solution:
4.16 (3.9596)
Exercise:
Problem: (1.07)(13.89)
Estimation by Clustering ([link])
For problems 71-90, estimate each value using the method of clustering.
After you have made an estimate, find the exact value. Compare the exact
and estimated values. Results may vary.
Exercise:
Problem: 38 + 51+ 41+ 48
Solution:
2(40) + 2(50) = 180 (178)
Exercise:
Problem: 19 + 73 + 23+ 71
Exercise:
Problem: 27 + 62 + 59 + 31
Solution:
2(30) + 2(60) = 180 (179)
Exercise:
Problem: 18 + 73 + 69 + 19
Exercise:
Problem: 83 + 49 + 79 + 52
Solution:
2(80) + 2(50) = 260 (263)
Exercise:
Problem: 67 + 71+ 84+ 81
Exercise:
Problem: 16 + 13 + 24+ 26
Solution:
3(20) + 1(10) = 70 (79)
Exercise:
Problem: 34 + 56 + 36+ 55
Exercise:
Problem: 14+ 17+ 83+ 87
Solution:
2(15) + 2(80) = 190 (201)
Exercise:
Problem: 93 + 108 + 96 + 111
Exercise:
Problem: 18 + 20 + 31 + 29 + 24+ 38
Solution:
3(20) + 2(30) + 40 = 160 (160)
Exercise:
Problem: 32 + 27 + 48+ 51+ 72+ 69
Exercise:
Problem: 64+ 17+ 27+ 59+ 31+ 21
Solution:
2(60) + 2(20) + 2(30) = 220 (219)
Exercise:
Problem: 81 + 41 + 92 + 38+ 88+ 80
Exercise:
Problem: 87 + 22 + 91
Solution:
2(90) + 20 = 200 (200)
Exercise:
Problem: 44 + 38 + 87
Exercise:
Problem: 19 + 18 + 39 + 22 + 42
Solution:
3(20) + 2(40) = 140 (140)
Exercise:
Problem: 31 + 28 + 49 + 29
Exercise:
Problem: 88 + 86 + 27+ 914+ 29
Solution:
3(90) + 2(30) = 330 (321)
Exercise:
Problem: 57 + 62+ 18+ 23+ 61+ 21
Mental Arithmetic- Using the Distributive Property ((link])
For problems 91-110, compute each product using the distributive property.
Exercise:
Problem: 15 - 33
Solution:
15(30 + 3) = 450 + 45 = 495
Exercise:
Problem: 15 - 42
Exercise:
Problem: 35 - 36
Solution:
35(40 — 4) = 1400 — 140 = 1,260
Exercise:
Problem: 35 - 28
Exercise:
Problem: 85 - 23
Solution:
85(20 + 3) = 1,700 + 225 = 1,955
Exercise:
Problem: 95- 11
Exercise:
Problem: 30 - 14
Solution:
30(10 + 4) = 300 + 120 = 420
Exercise:
Problem: 60 - 18
Exercise:
Problem: 75 - 23
Solution:
75(20 +3) = 1,500 + 225 = 1,725
Exercise:
Problem: 65 - 31
Exercise:
Problem: 17 - 15
Solution:
15(20 — 3) = 300 — 45 = 255
Exercise:
Problem: 38 - 25
Exercise:
Problem: 14 - 65
Solution:
65(10 + 4) = 650 + 260 = 910
Exercise:
Problem: 19 - 85
Exercise:
Problem: 42 - 60
Solution:
60(40 + 2) = 2,400 + 120 = 2,520
Exercise:
Problem: 81 - 40
Exercise:
Problem: 15 - 105
Solution:
15(100 a 5) = 1,500 + 75 = 1,575
Exercise:
Problem: 35 - 202
Exercise:
Problem: 45 - 306
Solution:
45(300 + 6) = 13,500 + 270 = 13,770
Exercise:
Problem: 85 - 97
Estimation by Rounding Fractions ([{link])
For problems 111-125, estimate each sum using the method of rounding
fractions. After you have made an estimate, find the exact value. Compare
the exact and estimated values. Results may vary.
Exercise:
~3 , 5
Problem: 3 + r
Solution:
1 2 ep 5
7 b= be
Exercise:
se
Problem: 716 + of
Exercise:
Problem: 1 + 30
Solution:
iE Ve Bho a
2 -+ a 1 ( or
Exercise:
. 19
Problem: ae ol op
Exercise:
Problem: 22 + me
Solution:
a +4 = 4(t0)
Exercise:
: 7
Problem: eee
Exercise:
Problem: =>
Solution:
Rial (or i.)
Exercise:
. 5 1
Problem: > + 35
Exercise:
<ignd 3
Problem: 2+ + 6<
Solution:
23+64=91(92)
Exercise:
Problem: 43 + 85-
Exercise:
‘ 5) 22
Problem: 115, + 74
Solution:
: 1 3 (4923
lf +75 = 183 (1823)
Exercise:
Problem: 1432 + 25
Exercise:
Problem: 65, + 2+. +8=
Solution:
6+2+84 = 164(161)
Exercise:
257 1 5
Problem: oF + re + 125
Exercise:
: 1 15 19
Problem: 10 + 6 ee 8 a
Solution:
1 ste 3 27
105 +7+84 = 254 (2547)
Proficiency Exam
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is a proficiency exam to the chapter
"Techniques of Estimation." Each problem is accompanied with a reference
link pointing back to the module that discusses the type of problem
demonstrated in the question. The problems in this exam are accompanied
by solutions.
Proficiency Exam
For problems 1 - 16, estimate each value. After you have made an estimate,
find the exact value. Results may vary.
Exercise:
Problem: ({link]) 3,716 + 6,789
Solution:
10,500 (10,505)
Exercise:
Problem: ((link]) 8,821 + 9,217
Solution:
18,000 (18,038)
Exercise:
Problem: ((link]) 7,316 — 2,305
Solution:
5,000 (5,011)
Exercise:
Problem: ([link]) 110,812 — 83,406
Solution:
28,000 (27,406)
Exercise:
Problem: ({link|) 82 - 38
Solution:
3,200 (3,116)
Exercise:
Problem: ({link]) 51 - 92
Solution:
4,500 (4,692)
Exercise:
Problem: ({link]) 48 - 6,012
Solution:
300,000 (288,576)
Exercise:
Problem: ((link]) 238 + 17
Solution:
12 (14)
Exercise:
Problem: ({link]) 2,660 + 28
Solution:
90 (95)
Exercise:
Problem: ((link]) 43.06 + 37.94
Solution:
81 (81.00)
Exercise:
Problem: ((link]) 307.006 + 198.0005
Solution:
505 (505.0065)
Exercise:
Problem: ((link]) (47.2)(92.8)
Solution:
4,371 (4,380.16)
Exercise:
Problem: ({link]) 58 + 91 + 61 + 88
Solution:
2(60) + 2(90) = 300 (298)
Exercise:
Problem: ({link]) 43 + 39 + 89 + 92
Solution:
2(40) + 2(90) = 260 (263)
Exercise:
Problem: ({link]) 81 + 78 + 27 + 79
Solution:
30 + 3(80) = 270 (265)
Exercise:
Problem: ([link]) 804 + 612 + 801 + 795 + 606
Solution:
3(800) + 2(600) = 3,600 (3,618)
For problems 17-21, use the distributive property to obtain the exact result.
Exercise:
Problem: ({link]) 25 - 14
Solution:
25(10 + 4) = 250 + 100 = 350
Exercise:
Problem: ({link]) 15 - 83
Solution:
15(80 + 3) = 1,200 + 45 = 1,245
Exercise:
Problem: ([link]) 65 - 98
Solution:
65(100 — 2) = 6,500 — 130 = 6,370
Exercise:
Problem: ({link]) 80 - 107
Solution:
80(100 + 7) = 8,000 + 560 = 8,560
Exercise:
Problem: ({link]) 400 - 215
Solution:
400(200 + 15) = 80,000 + 6,000 = 86,000
For problems 22-25, estimate each value. After you have made an estimate,
find the exact value. Results may vary.
Exercise:
Problem: ({link]) — a ~
Solution:
1+ 5 =15(156)
Exercise:
Problem: ({link] ) 35 + oT - 30
Solution:
1 1 47
04444-1114)
Exercise:
Problem: ({link]) 8 + 144
Solution:
1 _ 1 31
84 4 14 = 225 (2231)
Exercise:
Problem: ({link]) 5= + 13g + 675
Solution:
1 1 Ie = 1 1
By tle +65 = 135135)
Objectives
This module contains the learning objectives for the chapter "Measurement
and Geometry" from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, jr.
After completing this chapter, you should
Measurement and the United States System (({link])
e know what the word measurement means
e be familiar with United States system of measurement
e be able to convert from one unit of measure in the United States
system to another unit of measure
The Metric System of Measurement ({link])
e be more familiar with some of the advantages of the base ten number
system
e know the prefixes of the metric measures
e be familiar with the metric system of measurement
e be able to convert from one unit of measure in the metric system to
another unit of measure
Simplification of Denominate Numbers ((link})
e be able to convert an unsimplified unit of measure to a simplified unit
of measure
e be able to add and subtract denominate numbers
e be able to multiply and divide a denominate number by a whole
number
Perimeter and Circumference of Geometric Figures ({link]})
¢ know what a polygon is
e know what perimeter is and how to find it
e know what the circumference, diameter, and radius of a circle is and
how to find each one
e know the meaning of the symbol and its approximating value
e know what a formula is and four versions of the circumference
formula of a circle
Area and Volume of Geometric Figures and Objects ({link])
e know the meaning and notation for area
e know the area formulas for some common geometric figures
¢ be able to find the areas of some common geometric figures
e know the meaning and notation for volume
e know the volume formulas for some common geometric objects
¢ be able to find the volume of some common geometric objects
Measurement and the United States System
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module
discusses the United States System of measurement. By the end of the module students should know what the
word measurement means, be familiar with United States system of measurement and be able to convert from one
unit of measure in the United States system to another unit of measure.
Section Overview
e Measurement
e The United States System of Measurement
e Conversions in the United States System
Measurement
There are two major systems of measurement in use today. They are the United States system and the metric
system. Before we describe these systems, let's gain a clear understanding of the concept of measurement.
Measurement
Measurement is comparison to some standard.
Standard Unit of Measure
The concept of measurement is based on the idea of direct comparison. This means that measurement is the result
of the comparison of two quantities. The quantity that is used for comparison is called the standard unit of
measure.
Over the years, standards have changed. Quite some time in the past, the standard unit of measure was determined
by a king. For example,
1 inch was the distance between the tip of the thumb and the knuckle of the king.
1 inch was also the length of 16 barley grains placed end to end.
Today, standard units of measure rarely change. Standard units of measure are the responsibility of the Bureau of
Standards in Washington D.C.
Some desirable properties of a standard are the following:
1. Accessibility. We should have access to the standard so we can make comparisons.
2. Invariance. We should be confident that the standard is not subject to change.
3. Reproducibility. We should be able to reproduce the standard so that measurements are convenient and
accessible to many people.
The United States System of Measurement
Some of the common units (along with their abbreviations) for the United States system of measurement are listed
in the following table.
Unit Conversion Table
Length 1 foot (ft) = 12 inches (in.)
1 yard (yd) = 3 feet (ft)
1 mile (mi) = 5,280 feet
: 1 pound (1b) =16 ounces (0z)
Went 1 ton (T) = 2,000 pounds
1 tablespoon (tbsp) = 3 teaspoons (tsp)
1 fluid ounce (fl oz) = 2 tablespoons
1 cup (c) = 8 fluid ounces
1 pint (pt) = 2 cups
1 quart (qt) = 2 pints
1 gallon (gal) = 4 quarts
Liquid Volume
1 minute (min) = 60 seconds (sec)
1 hour ( hr) = 60 minutes
1 day (da) = 24 hours
1 week (wk) = 7 days
Time
Conversions in the United States System
It is often convenient or necessary to convert from one unit of measure to another. For example, it may be
convenient to convert a measurement of length that is given in feet to one that is given in inches. Such
conversions can be made using unit fractions.
Unit Fraction
A unit fraction is a fraction with a value of 1.
Unit fractions are formed by using two equal measurements. One measurement is placed in the numerator of the
fraction, and the other in the denominator. Placement depends on the desired conversion.
Placement of Units
Place the unit being converted to in the numerator.
Place the unit being converted from in the denominator.
For example,
Equal Measurements Unit Fraction
_ 49: 1ft . 12in.
1ft = 12in. Ol ae
lpt = 16 fl oz wt aa
—_ 7da lwk
lwk = 7da iwk OF 7da
Sample Set A
Make the following conversions. If a fraction occurs, convert it to a decimal rounded to two decimal places.
Example:
Convert 11 yards to feet.
Looking it we unit conversion table under length, we see that lyd = 3 ft. There are two corresponding unit
and ok = . Which one should we use? Look to see which unit we wish to convert to. Choose the unit
? : Sra
fraction with this ee in the numerator. We will choose 2 oo since this unit fraction has feet in the numerator.
Now, multiply 11 yd by the unit fraction. Notice that since the unit fraction has the value of 1, multiplying by it
does not change the value of 11 yd.
Lilyel = Se a Divide out common units.
fractions
= ane dite (Units can be added, subtracted, multiplied, and divided, just as numbers can.)
33ft
Thus, llyd = = Oat,
Example:
Convert 36 fl oz to pints.
Looking in the unit conversion table under liquid volume, we see that 1 pt = 16 fl oz. Since we are to convert to
pints, we will construct a unit fraction with pints in the numerator.
Ipt ere 3
36floz = pH on. : = = Divide out common units.
3660 pt
a - 16 floz
36-1 pt
16
t
= aye Reduce.
= tpt Convert to decimals: 4 = 2.25.
Thus, 36 fl oz = 2.25 pt.
Example:
Convert 2,016 hr to weeks.
Looking in the unit conversion table under time, we see that lwk = 7da and that 1da = 24 hr. To convert from
hours to weeks, we must first convert from hours to days and then from days to weeks. We need two unit
fractions.
The unit fraction needed for converting from hours to days i
1 wk
days to weeks is +3..
_—-2,016hr lda lwk ware j
2,016hr = i Sa ae Divide out common units.
_ 2,016b lds wk
1 WM pe 7h
_-2,016-1wk
= 7 — Reduce.
12wk
Thus, 2 016 hr = 12 wk.
Practice Set A
Make the following conversions. If a fraction occurs, convert it to a decimal rounded to two decimal places.
Exercise:
Problem: Convert 18 ft to yards.
Solution:
6 yd
Exercise:
Problem: Convert 2 mi to feet.
Solution:
10,560 ft
Exercise:
Problem: Convert 26 ft to yards.
Solution:
8.67 yd
Exercise:
Problem: Convert 9 qt to pints.
Solution:
18 pt
Exercise:
Problem: Convert 52 min to hours.
Solution:
0.87 hr
Exercise:
Problem: Convert 412 hr to weeks.
Solution:
2.45 wk
Exercises
Make each conversion using unit fractions. If fractions occur, convert them to decimals rounded to two decimal
places.
Exercise:
Problem: 14 yd to feet
Solution:
42 feet
Exercise:
Problem: 3 mi to yards
Exercise:
Problem: 8 mi to inches
Solution:
506,880 inches
Exercise:
Problem: 2 mi to inches
Exercise:
Problem: 18 in. to feet
Solution:
1.5 feet
Exercise:
Problem: 84 in. to yards
Exercise:
Problem: 5 in. to yards
Solution:
0.14 yard
Exercise:
Problem: 106 ft to miles
Exercise:
Problem: 62 in. to miles
Solution:
0.00 miles (to two decimal places)
Exercise:
Problem: 0.4 in. to yards
Exercise:
Problem: 3 qt to pints
Solution:
6 pints
Exercise:
Problem: 5 |b to ounces
Exercise:
Problem: 6 T to ounces
Solution:
192,000 ounces
Exercise:
Problem: 4 oz to pounds
Exercise:
Problem: 15,000 oz to pounds
Solution:
937.5 pounds
Exercise:
Problem: 15,000 oz to tons
Exercise:
Problem: 9 tbsp to teaspoons
Solution:
27 teaspoons
Exercise:
Problem: 3 c to tablespoons
Exercise:
Problem: 5 pt to fluid ounces
Solution:
80 fluid ounces
Exercise:
Problem: 16 tsp to cups
Exercise:
Problem: 5 fl oz to quarts
Solution:
0.16 quart
Exercise:
Problem: 3 qt to gallons
Exercise:
Problem: 5 pt to teaspoons
Solution:
480 teaspoons
Exercise:
Problem: 3 qt to tablespoons
Exercise:
Problem: 18 min to seconds
Solution:
1,080 seconds
Exercise:
Problem: 4 days to hours
Exercise:
Problem: 3 hr to days
Solution:
+ = 0.125 day
Exercise:
Problem: + hr to days
Exercise:
Problem: + da to weeks
Solution:
=; = 0.0714 week
Exercise:
Problem: 37 wk to seconds
Exercises for Review
Exercise:
Problem: ((link]) Specify the digits by which 23,840 is divisible.
Solution:
1,2,4,5,8
Exercise:
Problem: ((link]) Find 25 of 53 of 72.
Exercise:
Problem: ((link]) Convert 0.34 to a fraction.
Solution:
il
30
Exercise:
Problem: ((link]) Use the clustering method to estimate the sum: 53 + 82 + 79 + 49.
Exercise:
Problem: ((link]) Use the distributive property to compute the product: 60 - 46.
Solution:
60(50 — 4) = 3,000 — 240 = 2,760
The Metric System of Measurement
This module is from Fundamentals of Mathematics by Denny Burzynski and
Wade Ellis, Jr. This module discusses the Metric System of measurement. By
the end of the module students should be more familiar with some of the
advantages of the base ten number system, know the prefixes of the metric
measures, be familiar with the metric system of measurement and be able to
convert from one unit of measure in the metric system to another unit of
measure
Section Overview
e The Advantages of the Base Ten Number System
e Prefixes
e Conversion from One Unit to Another Unit
e¢ Conversion Table
The Advantages of the Base Ten Number System
The metric system of measurement takes advantage of our base ten number sys-
tem. The advantage of the metric system over the United States system is that in
the metric system it is possible to convert from one unit of measure to another
simply by multiplying or dividing the given number by a power of 10. This
means we can make a conversion simply by moving the decimal point to the
right or the left.
Prefixes
Common units of measure in the metric system are the meter (for length), the
liter (for volume), and the gram (for mass). To each of the units can be attached
a prefix. The metric prefixes along with their meaning are listed below.
Metric Prefixes
¢ kilothousand
e decitenth
e hectohundred
e centihundredth
e dekaten
e millithousandth
For example, if length is being measured,
1 kilometer is equivalent to 1000 meters.
1 centimeter is equivalent to one hundredth of a meter.
1 millimeter is equivalent to one thousandth of a meter.
Conversion from One Unit to Another Unit
Let's note three characteristics of the metric system that occur in the metric table
of measurements.
1. In each category, the prefixes are the same.
2. We can move from a larger to a smaller unit of measure by moving the
decimal point to the right.
3. We can move from a smaller to a larger unit of measure by moving the
decimal point to the left.
The following table provides a summary of the relationship between the basic
unit of measure (meter, gram, liter) and each prefix, and how many places the
decimal point is moved and in what direction.
kilo hecto deka unit deci centi milli
Basic Unit to Prefix Move the Decimal Point
unit to deka 1 to 10 1 place to the left
unit to hector 1 to 100 2 places to the left
unit to kilo 1 to 1,000 3 places to the left
unit to deci 1to0.1 1 place to the right
unit to centi 1 to 0.01 2 places to the right
unit to milli 1 to 0.001 3 places to the right
Conversion Table
Listed below, in the unit conversion table, are some of the common metric units
of measure.
Unit Conversion Table
1 kilometer (km) = 1,000 meters ( m) 1,000 x 1m
1 hectometer (hm) = 100 meters 100 x 1m
1 dekameter ( dam) = 10 meters 10 x Im
Length 1 meter (m) 1 x 1m
1 decimeter (dm) = => meter 1x 1m
1 centimeter (cm) = 7, meter 01 x 1m
1 millimeter (mm) = T0008 meter .001 x Im
Mass 1 kilogram (kg) = 1,000 grams ( g) 1,000 x 1g
1 hectogram (hg) = 100 grams 100 x 1g
1 dekagram ( dag) = 10 grams 10 x lg
1 gram (g) 1x lg
1 decigram (dg) = 7 gram 1xilg
1 centigram (cg) = 7 gram 01 x 1g
1 milligram (mg) = [000 gram .001 x 1g
1 kiloliter (kL) = 1,000 liters (LZ) 1,000 x 1L
1 hectoliter ( hL) = 100 liters 100 x 1L
1 dekaliter (daL) = 10 liters 10 x 1L
Voluinie 1 liter (L) 1x 1L
1 deciliter (dL) = 35 liter 1x 1L
1 centiliter (cL) = 7 liter 01 x 1L
1 milliliter (mL) = {ooo liter 001 x 1L
Time Same as the United States system
Distinction Between Mass and Weight
There is a distinction between mass and weight. The weight of a body is related
to gravity whereas the mass of a body is not. For example, your weight on the
earth is different than it is on the moon, but your mass is the same in both
places. Mass is a measure of a body's resistance to motion. The more massive a
body, the more resistant it is to motion. Also, more massive bodies weigh more
than less massive bodies.
Converting Metric Units
To convert from one metric unit to another metric unit:
1. Determine the location of the original number on the metric scale (pictured
in each of the following examples).
2. Move the decimal point of the original number in the same direction and
same number of places as is necessary to move to the metric unit you wish
to go to.
We can also convert from one metric unit to another using unit fractions. Both
methods are shown in [link] of [link].
Sample Set A
Example:
Convert 3 kilograms to grams.
a. 3 kg can be written as 3.0 kg. Then,
pae Oe Oe gs dg cg, mg,
“4 —
1 2 3
places to the right
3.0 kg = 3.000. g
123
Thus, 3kg = 3,000 g.
b. We can also use unit fractions to make this conversion.
Since we are converting to grams, and 1,000 g=1 kg, we choose the
. 1,000 g_. ae
unit fraction — aa since grams is in the numerator.
3 kg = 3kg- +08
= 3 yf tas
Example:
Convert 67.2 hectoliters to milliliters.
L—_ _—. = => FL
1 2 3 4 5
places to the right
67.2 hL = 67 20000, mL
12345
Thus, 67.2 hL = 6,720,000 mL.
Example:
Convert 100.07 centimeters to meters.
km , hm , dam, m dm, cm mm
places to the left
100.07 cm = 1,0007 m
2
RAY
1
Thus, 100.07 cm = 1.0007 m.
Example:
Convert 0.16 milligrams to grams.
ke, he , de, gs | de ce mg,
places to the left
0.16 mg = 0,000 16 g
321
Thus, 0.16 mg = 0.00016 g.
Practice Set A
Exercise:
Problem: Convert 411 kilograms to grams.
Solution:
411,000 g
Exercise:
Problem: Convert 5.626 liters to centiliters.
Solution:
962.6 cL
Exercise:
Problem: Convert 80 milliliters to kiloliters.
Solution:
0.00008 kL
Exercise:
Problem: Convert 150 milligrams to centigrams.
Solution:
15 cg
Exercise:
Problem: Convert 2.5 centimeters to meters.
Solution:
0.025 m
Exercises
Make each conversion.
Exercise:
Problem: 87 m to cm
Solution:
8,700 cm
Exercise:
Problem
Exercise:
Problem
> 905 L to mL
: 16,005 mg to g
Solution:
16.005 g
Exercise:
Problem
Exercise:
Problem
: 48.66 L to dL
: 11.161 kL toL
Solution:
11,161 L
Exercise:
Problem
Exercise:
: 521.85 cm to mm
Problem
: 1.26 dag to dg
Solution:
126 dg
Exercise:
Problem
Exercise:
Problem
: 99.04 dam to cm
: 0.51 kL to daL
Solution:
5.1 daL
Exercise:
Problem
Exercise:
Problem
: 0.17 kL to daL
: 0.05 mto dm
Solution:
0.5 dm
Exercise:
Problem
Exercise:
Problem
: 0.001 km to mm
: 8.106 hg to cg
Solution:
81,060 cg
Exercise:
Problem:
Exercise:
Problem:
Solution:
0.03 m
Exercise:
Problem:
Exercise:
Problem:
Solution:
4,000 mg
Exercise:
Problem:
Exercise:
Problem:
Solution:
17.0186 kL to mL
3cmtom
9mmtom
4gtomg
2 110:kL
6 kg to mg
6,000,000 mg
Exercise:
Problem:
7 daL to mL
Exercises for Review
Exercise:
vee a oe
Problem: ((link]) Find the value of 3 — = + 7.
Solution:
Exercise:
- (Ti; a Oe
Problem: ((link]) Solve the proportion: —- = <7.
Exercise:
Problem:
({link]) Use the method of rounding to estimate the sum: 8,226 + 4,118.
Solution:
12,300 (12,344)
Exercise:
Problem:
({link]) Use the clustering method to estimate the sum:
87 +121 + 118 + 91 + 92.
Exercise:
Problem: ((link]) Convert 3 in. to yd.
Solution:
0.083 yard
Simplification of Denominate Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to simplify denominate
numbers. By the end of the module students should be able to convert an
unsimplified unit of measure to a simplified unit of measure, be able to add
and subtract denominate numbers and be able to multiply and divide a
denominate number by a whole number.
Section Overview
e Converting to Multiple Units
e Adding and Subtracting Denominate Numbers
e Multiplying a Denominate Number by a Whole Number
e Dividing a Denominate Number by a Whole Number
Converting to Multiple Units
Denominate Numbers
Numbers that have units of measure associated with them are called
denominate numbers. It is often convenient, or even necessary, to simplify
a denominate number.
Simplified Denominate Number
A denominate number is simplified when the number of standard units of
measure associated With it does not exceed the next higher type of unit.
The denominate number 55 min is simplified since it is smaller than the
next higher type of unit, 1 hr. The denominate number 65 min is not
simplified since it is not smaller than the next higher type of unit, 1 hr. The
denominate number 65 min can be simplified to 1 hr 5 min. The
denominate number 1 hr 5 min is simplified since the next higher type of
unit is day, and 1 hr does not exceed 1 day.
Sample Set A
Example:
Simplify 19 in.
Sinced2: in, = lttvand 19 = 12-5 7,
19in. = 12in.4+7in.
= Ltp= 7 in:
ante nae
Example:
Simplify 4 gal 5 qt.
Since 4 qt = 1 gal, and5 = 4+1,
Agaldqt = 4gal+4qt+ 1lqt
— 4eal- eal- lat
= 5 gal+1qt
Sore alia
Example:
Simplify 2 hr 75 min.
Since 60 min = 1 hr, and 75 = 60+ 15,
2hr75min = 2hr+60 min+ 15min
= 2hr+1hr+15 min
= 3hr+15min
3 hr 15 min
Example:
Simplify 43 fl oz.
Since 8 fl oz = 1 c (1 cup), and 43 = 8 = 5R3,
43 floz = 40 floz+3floz
5-8 floz-+ 3 fl oz
= 5-1lc+3floz
5c+3 fl oz
But, 2¢ = 1 pt ands — 2 — 2R 1, So;
5c+3floz = 2-2c+1c+3floz
==) og leap aril oy
= 2pt+1c+3floz
But, 2 pt — lat,.so
2pt+1c+3floz=1qt1c3floz
Practice Set A
Simplify each denominate number. Refer to the conversion tables given in
[link], if necessary.
Exercise:
Problem: 18 in.
Solution:
1 ft 6 in.
Exercise:
Problem: 8 gal 9 gt
Solution:
10 gal 1 qt
Exercise:
Problem: 5 hr 80 min
Solution:
6 hr 20 min
Exercise:
Problem: 8 wk 11 da
Solution:
9 wk 4 da
Exercise:
Problem: 86 da
Solution:
12 wk 2 da
Adding and Subtracting Denominate Numbers
Adding and Subtracting Denominate Numbers
Denominate numbers can be added or subtracted by:
1. writing the numbers vertically so that the like units appear in the same
column.
2. adding or subtracting the number parts, carrying along the unit.
3. simplifying the sum or difference.
Sample Set B
Example:
Add 6 ft 8 in. to 2 ft 9 in.
6 ft 8 in.
+2 ft 9 in.
8ft17in. Simplify this denominate number.
Since: 24am. = Et:
Sit 4-12 in, + Sins = Sit 1 it 5 in,
= 9ft+5in.
= Ohio ine
Example:
Subtract 5 da 3 hr from 8 da 11 hr.
8 da 11 hr
—5ida 3hr
3da 8hr
Example:
Subtract 3 lb 14 oz from 5 Ib 3 oz.
5lb 302
—3 lb 14 oz
We cannot directly subtract 14 oz from 3 oz, so we must borrow 16 oz
from the pounds.
5lb30z = 51b+302
= 4)b-- llb +302
4lb+160z+30z (Since 1 lb = 16 0z.)
— 2 1b 19107
4 lb 19 oz
4 lb 19 oz
—3 lb 14 oz
1lb 5oz
Example:
Subtract 4 da 9 hr 21 min from 7 da 10 min.
7 da 0 hr 10 min
B e
re tn ee a orrow 1 da from the 7 da
6 da 24 hr 10 min
B lhrf he 24 hr.
ete on orrow 1 hr from the ih
6 da 23 hr 70 min
—4da 9 hr 21 min
2 da 14 hr 49 min
Practice Set B
Perform each operation. Simplify when possible.
Exercise:
Problem: Add 4 gal 3 gt to 1 gal 2 qt.
Solution:
6 gal 1 gt
Exercise:
Problem: Add 9 hr 48 min to 4 hr 26 min.
Solution:
14 hr 14 min
Exercise:
Problem: Subtract 2 ft 5 in. from 8 ft 7 in.
Solution:
6 ft 2in.
Exercise:
Problem: Subtract 15 km 460 m from 27 km 800 m.
Solution:
12 km 340 m
Exercise:
Problem: Subtract 8 min 35 sec from 12 min 10 sec.
Solution:
3 min 35 sec
Exercise:
Problem: Add 4 yd 2 ft 7 in. to 9 yd 2 ft 8 in.
Solution:
14 yd 2 ft 3 in
Exercise:
Problem: Subtract 11 min 55 sec from 25 min 8 sec.
Solution:
13 min 13 sec
Multiplying a Denominate Number by a Whole Number
Let's examine the repeated sum
4ft9in. + 4ft9in. + 4ft 9in. = 12 ft 27 in.
3 times
Recalling that multiplication is a description of repeated addition, by the
distributive property we have
3(4ft9in.) = 3(4ft+9 in.)
= 9° 4Tt-+ 3 = 9:in.
= 12ft+ 27 in. Now; 27 in.= 2 ft 3 in,
= 12ft+2ft+3 in.
= 14ft+ 3 in.
14 ft 3 in.
From these observations, we can suggest the following rule.
Multiplying a Denominate Number by a Whole Number
To multiply a denominate number by a whole number, multiply the number
part of each unit by the whole number and affix the unit to this product.
Sample Set C
Perform the following multiplications. Simplify if necessary.
Example:
6-Qit4in.) = 6-21t+ 6-4 in.
= 12ft+ 24in.
Since 3 ft = 1 yd and 12 in. = 1 ft,
12ft+24in. = 4yd+2 ft
avai at
Example:
8-(5hr 21min 55sec) = 8-5hr+8-21min+ 8-55 sec
= 40 hr+ 168 min + 440sec
= 40 hr+ 168 min + 7 min + 20 sec
= AQhr+175 min + 20 sec
= 40 hr+ 2 hr+ 55 min + 20 sec
= 42 hr+ 55 min + 20 sec
= 24hr + 18hr + 55 min + 20 sec
= lda+18 hr+ 55 min + 20 sec
= 1da18hr 55 min 20 sec
Practice Set C
Perform the following multiplications. Simplify.
Exercise:
Problem: 2 - (10 min)
Solution:
20 min
Exercise:
Problem: 5 - (3 qt)
Solution:
15 qt = 3 gal 3 qt
Exercise:
Problem: 4 - (5ft 8 in.)
Solution:
201.32. mo= 7 yd LTt:3 in;
Exercise:
Problem: 10 - (2hr 15 min 40 sec)
Solution:
20 hr 150 min 400 sec = 22 hr 36 min 40 sec
Dividing a Denominate Number by a Whole Number
Dividing a Denominate Number by a Whole Number
To divide a denominate number by a whole number, divide the number part
of each unit by the whole number beginning with the largest unit. Affix the
unit to this quotient. Carry any remainder to the next unit.
Sample Set D
Perform the following divisions. Simplify if necessary.
Example:
(12 min 40 sec) + 4
3 min 10 sec
4) 12 min 40 sec
12 min
40 sec
40 sec
0
Thus (12 min 40 sec) + 4 = 3min 10 sec
Example:
(5 yd2 ft 9in.) +3
1 yd 2 ft 11 in.
3)5 yd2 ft 9 in.
Convert to feet: 2 yd2ft = 8ft
Convert toinches:2ft9in . = 33in
Thus (5 yd 2 ft 9 in.) + 3 = lyd 2 ft 11 in.
Practice Set D
Perform the following divisions. Simplify if necessary.
Exercise:
Problem: (18 hr 36 min) + 9
Solution:
2 hr 4 min
Exercise:
Problem: (34 hr 8 min.) + 8
Solution:
4 hr 16 min
Exercise:
Problem: (13 yd 7 in.) + 5
Solution:
2 yd 1 ft 11in
Exercise:
Problem: (47 gal 2 qt 1 pt) +3
Solution:
15 gal 3 qt 1 pt
Exercises
For the following 15 problems, simplify the denominate numbers.
Exercise:
Problem: 16 in.
Solution:
1 foot 4 inches
Exercise:
Problem: 19 ft
Exercise:
Problem: 85 min
Solution:
1 hour 25 minutes
Exercise:
Problem: 90 min
Exercise:
Problem: 17 da
Solution:
2 weeks 3 days
Exercise:
Problem: 25 oz
Exercise:
Problem: 240 oz
Solution:
15 pounds
Exercise:
Problem: 3,500 |b
Exercise:
Problem: 26 gt
Solution:
6 gallons 2 quarts
Exercise:
Problem: 300 sec
Exercise:
Problem: 135 oz
Solution:
8 pounds 7 ounces
Exercise:
Problem: 14 tsp
Exercise:
Problem: 18 pt
Solution:
2 gallons 1 quart
Exercise:
Problem: 3,500 m
Exercise:
Problem: 16,300 mL
Solution:
16 liters 300 milliliters (or 1daL 6 L 3dL)
For the following 15 problems, perform the indicated operations and
simplify the answers if possible.
Exercise:
Problem: Add 6 min 12 sec to 5 min 15 sec.
Exercise:
Problem: Add 14 da 6 hr to 1 da 5 hr.
Solution:
15 days 11 hours
Exercise:
Problem: Add 9 gal 3 qt to 2 gal 3 qt.
Exercise:
Problem: Add 16 lb 10 oz to 42 lb 15 oz.
Solution:
59 pounds 9 ounces
Exercise:
Problem: Subtract 3 gal 1 gt from 8 gal 3 at.
Exercise:
Problem: Subtract 3 ft 10 in. from 5 ft 8 in.
Solution:
1 foot 10 inches
Exercise:
Problem: Subtract 5 lb 9 oz from 12 lb 5 oz.
Exercise:
Problem: Subtract 10 hr 10 min from 11 hr 28 min.
Solution:
1 hour 18 minutes
Exercise:
Problem: Add 3 fl oz 1 tbsp 2 tsp to 5 fl oz 1 tbsp 2 tsp.
Exercise:
Problem: Add 4 da 7 hr 12 min to 1 da 8 hr 53 min.
Solution:
5 days 16 hours 5 minutes
Exercise:
Problem: Subtract 5 hr 21 sec from 11 hr 2 min 14 sec.
Exercise:
Problem: Subtract 6 T 1,300 lb 10 oz from 8 T 400 lb 10 oz.
Solution:
1 ton 1,100 pounds (or 1T 1,100 Ib)
Exercise:
Problem: Subtract 15 mi 10 in. from 27 mi 800 ft 7 in.
Exercise:
Problem:
Subtract 3 wk 5 da 50 min 12 sec from 5 wk 6 da 20 min 5 sec.
Solution:
2 weeks 23 hours 29 minutes 53 seconds
Exercise:
Problem: Subtract 3 gal 3 gt 1 pt 1 oz from 10 gal 2 qt 2 oz.
Exercises for Review
Exercise:
Problem: ({link]) Find the value: (2) ° =P ve
Solution:
1
Exercise:
Problem: ((link]) Find the sum: 8 + 6%.
Exercise:
Problem: ({link]) Convert 2.05 to a fraction.
Solution:
14
2 275
Exercise:
Problem:
({link]) An acid solution is composed of 3 parts acid to 7 parts water.
How many parts of acid are there in a solution that contains 126 parts
water?
Exercise:
Problem: ({link]) Convert 126 kg to grams.
Solution:
126,000 g
Perimeter and Circumference of Geometric Figures
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr.
This module discusses perimeter and circumference of geometric figures. By the end of the
module students should know what a polygon is, know what perimeter is and how to find it,
know what the circumference, diameter, and radius of a circle is and how to find each one,
know the meaning of the symbol z and its approximating value and know what a formula is
and four versions of the circumference formula of a circle.
Section Overview
e Polygons
e Perimeter
e Circumference/Diameter/Radius
e The Number 7
e Formulas
Polygons
We can make use of conversion skills with denominate numbers to make measurements of
geometric figures such as rectangles, triangles, and circles. To make these measurements we
need to be familiar with several definitions.
Polygon
A polygon is a closed plane (flat) figure whose sides are line segments (portions of straight
lines).
Polygons
Not polygons
SOOL
Perimeter
Perimeter
The perimeter of a polygon is the distance around the polygon.
To find the perimeter of a polygon, we simply add up the lengths of all the sides.
Sample Set A
Find the perimeter of each polygon.
Example:
Perimeter = 2cm+5cm+2cm+5cm
= iam
Example:
Ne) oe
ey, S Ny
y ee
*
3.1 mm \
9.2 cm
Perimeter = 3.1 mm
4.2 mm
4.3 mm
1.52 mm
5.4 mm
+9.2 mm
27.72 mm
Example:
2cm
zen]
= |
9em
L 12cm ;
lem} rs lcm
lem
Our first observation is that three of the dimensions are missing. However, we can determine
the missing measurements using the following process. Let A, B, and C represent the
missing measurements. Visualize
| $$$
Lf Cc
1
IA = 12m — 2m = 10m
B=9m-+ 1m — 2m = 8m
GO = 1 = iia = iho
Perimeter = 8m
10m
Practice Set A
Find the perimeter of each polygon.
Exercise:
Problem:
\— 9 ft
3 ft i,
; 8 ft
Solution:
20 ft
Exercise:
Problem:
5.8m
6.1m 6.3 m
8.6m
Solution:
26.8 m
Exercise:
Problem:
10.07 mi
3.11 mi 3.88 mi
4.54 mi
10.76 mi
4.92 mi
12.61 mi
Solution:
49.89 mi
Circumference/Diameter/Radius
Circumference
The circumference of a circle is the distance around the circle.
Diameter
A diameter of a circle is any line segment that passes through the center of the circle and has
its endpoints on the circle.
Radius
A radius of a circle is any line segment having as its endpoints the center of the circle and a
point on the circle.
The radius is one half the diameter.
Diameter ,
The Number z
The symbol 7, read "pi," represents the nonterminating, nonrepeating decimal number
3.14159 ... . This number has been computed to millions of decimal places without the
appearance of a repeating block of digits.
For computational purposes, 7 is often approximated as 3.14. We will write 7 ~ 3.14 to
denote that 7 is approximately equal to 3.14. The symbol "*" means "approximately equal
to."
Formulas
To find the circumference of a circle, we need only know its diameter or radius. We then use
a formula for computing the circumference of the circle.
Formula
A formula is a rule or method for performing a task. In mathematics, a formula is a rule that
directs us in computations.
Formulas are usually composed of letters that represent important, but possibly unknown,
quantities.
If C,, d, and r represent, respectively, the circumference, diameter, and radius of a circle, then
the following two formulas give us directions for computing the circumference of the circle.
Circumference Formulas
1.C = ndorC & (3.14)d
2,C = 2arorC = 2(3.14)r
Sample Set B
Example:
Find the exact circumference of the circle.
Use the formula C' = zd.
C=7 in,
By commutativity of multiplication,
C= neat
C' = 7zin., exactly
This result is exact since 7 has not been approximated.
Example:
Find the approximate circumference of the circle.
Use the formula C' = zd.
C = (3.14)(6.2)
C + 19.648 mm
This result is approximate since 7 has been approximated by 3.14.
Example:
Find the approximate circumference of a circle with radius 18 inches.
Since we're given that the radius, r, is 18 in., we'll use the formula C' = 2zr.
C22 (SA Pietra)
C= 113.04 in.
Example:
Find the approximate perimeter of the figure.
We notice that we have two semicircles (half circles).
The larger radius is 6.2 cm.
The smaller radius is 6.2 cm - 2.0 cm = 4.2 cm.
The width of the bottom part of the rectangle is 2.0 cm.
Perimeter = 2.0cm
5.1 cm
2.0 cm
5.1 cm
(0.5) - (2) - (3.14) - (6.2 cm) Circumference of outer semicircle.
+ (0.5) - (2) -(3.14)- (4.2 cm) Circumference of inner semicircle
Perimeter ~ 2.0 cm
5.1 cm
BOs Cre
5.1 cm
19.468 cm
+13.188 cm
48.856 cm
Practice Set B
Exercise:
6.2 cm — 2.0 cm = 4.2 cm
The 0.5 appears because we want the
perimeter of only half a circle.
Problem: Find the exact circumference of the circle.
d= AD |
=e
——
Solution:
9.17 in.
Exercise:
Problem: Find the approximate circumference of the circle.
vA |
VA \
m \
= 1 3B ae
d = —_—— — |
co }
\ /
Solution:
5.652 mm
Exercise:
Problem: Find the approximate circumference of the circle with radius 20.1 m.
Solution:
126.228 m
Exercise:
Problem: Find the approximate outside perimeter of
| 16.2 mm
Solution:
41.634 mm
Exercises
Find each perimeter or approximate circumference. Use a = 3.14.
Exercise:
Problem:
8.6 cm
Solution:
21.8 cm
Exercise:
Problem:
8 mum
Exercise:
Problem:
Solution:
38.14 inches
Exercise:
Problem:
0.04 ft Tis
0.095 ft
Exercise:
Problem:
0.12 m
0.31 m
Solution:
0.86 m
Exercise:
Problem:
Exercise:
Problem:
(2
a
Solution:
87.92 m
Exercise:
Problem:
<
| N38 mm
\
%
*,
\
Exercise:
Problem:
Solution:
16.328 cm
Exercise:
Problem:
1.1 mm
Exercise:
Problem:
0.03 cm
———— —
Solution:
0.0771 cm
Exercise:
Problem:
5 in.
Exercise:
Problem:
Solution:
120.78 m
Exercise:
Problem:
4.1 in.
7.8 in.
Exercise:
Problem:
Solution:
21.71 inches
Exercise:
Problem:
18m
Exercise:
Problem:
Solution:
43.7 mm
Exercise:
Problem:
Exercise:
Problem:
Solution:
45.68 cm
Exercise:
Problem:
Exercises for Review
Exercise:
Problem: ((link]) Find the value of 24 : / 103.
Solution:
17 1
8.5 or = OF Ss
Exercise:
Problem: ((link]) Find the value of = + a + ae
Exercise:
Problem: ({link]) Convert t to a decimal.
Solution:
0.875
Exercise:
Problem:
([link]) What is the name given to a quantity that is used as a comparison to determine
the measure of another quantity?
Exercise:
Problem: ((link]) Add 42 min 26 sec to 53 min 40 sec and simplify the result.
Solution:
1 hour 36 minutes 6 seconds
Area and Volume of Geometric Figures and Objects
This module is from Fundamentals of Mathematics by Denny Burzynski and
Wade Ellis, Jr. This module discusses area and volume of geometric figures and
objects. By the end of the module students should know the meaning and
notation for area, know the area formulas for some common geometric figures,
be able to find the areas of some common geometric figures, know the meaning
and notation for volume, know the volume formulas for some common
geometric objects and be able to find the volume of some common geometric
objects.
Section Overview
¢ The Meaning and Notation for Area
e Area Formulas
e Finding Areas of Some Common Geometric Figures
e The Meaning and Notation for Volume
¢ Volume Formulas
e Finding Volumes of Some Common Geometric Objects
Quite often it is necessary to multiply one denominate number by another. To do
so, we multiply the number parts together and the unit parts together. For
example,
8in.-8in. = 8-8-in.-in.
64 in.”
4mm:-4mm-4mm = 4-4-4-mm-mm-mm
64 mm?
Sometimes the product of units has a physical meaning. In this section, we will
examine the meaning of the products (length unit)? and (length unit)?.
The Meaning and Notation for Area
The product (length unit) - (length unit) = (length unit)’, or, square length
unit (sq length unit), can be interpreted physically as the area of a surface.
Area
The area of a surface is the amount of square length units contained in the
surface.
For example, 3 sq in. means that 3 squares, 1 inch on each side, can be placed
precisely on some surface. (The squares may have to be cut and rearranged so
they match the shape of the surface.)
We will examine the area of the following geometric figures.
——— 6 = base ———+
Triangles
w = width
+ —— |= length ———>
Rectangles
——— b = base —__>
Parallelograms
b, = base 2
———>
| ;
| h= height
||
————— b, = base 1 —_—_——_>
Trapezoids
Circles
Area Formulas
We can determine the areas of these geometric figures using the following
formulas.
Figure Area Formula Statement
Area of a
triangle is
/\. Triangle Arp =+-b-h one half the
base times
the height.
Area of a
rectangle is
|] Rectangle Ap=l-w the length
times the
width.
Parallelogram Ap=b-h Area of a
hy parallelogram
is base times
the height.
aa Trapezoid At = $ -(b; + bg) -h
C) Circle Ac = nr?
Finding Areas of Some Common Geometric Figures
Sample Set A
Example:
Find the area of the triangle.
Aen
20 ft
Ai = -b-h
- 20-6 sq ft
0-6 sqft
= 60 sq ft
60 ft?
Area of a
trapezoid is
one half the
sum of the
two bases
times the
height.
Area of a
circle is 7
times the
square of the
radius.
The area of this triangle is 60 sq ft, which is often written as 60 ft.
Example:
Find the area of the rectangle.
is in.
4 ft 2 in.
Let's first convert 4 ft 2 in. to inches. Since we wish to convert to inches, we'll
use the unit fraction We uh since it has inches in the numerator. Then,
At, — Aft . l2in.
=e 1ft
_ 4X | in.
1 Le
= Asin.
Thus, 4 ft 2in. = 48 in. + 2 in. = 50 in.
Ar ae AKG
= 50 in. 1m:
= 400 sq in.
The area of this rectangle is 400 sq in.
Example:
Find the area of the parallelogram.
10.3 cm
LANs ee boa re
10.3 cm: 6.2 cm
= 63.86 sq cm
The area of this parallelogram is 63.86 sq cm.
Example:
Find the area of the trapezoid.
14.5 mm
| 14.1 mm
20.4 mm
Atrap = 7+ (b1 +b2)-h
= +- (14.5 mm, +, 20.4 mm) - (4.1 mm)
= +-(34.9 mm) - (4.1 mm)
= +- (143.09 sq mm)
= 71.545 sq mm
The area of this trapezoid is 71.545 sq mm.
Example:
Find the approximate area of the circle.
Vie ss Oa
= (3.14) - (16.8 ft)”
(3.14) - (282.24 sq ft)
~ 888.23 sq ft
The area of this circle is approximately 886.23 sq ft.
2
Practice Set A
Find the area of each of the following geometric figures.
Exercise:
Problem:
18 cm
Solution:
36 sq cm
Exercise:
Problem:
4.05 mm
9.26 mm
Solution:
37.503 sq mm
Exercise:
Problem:
ind
Solution:
13.26 sq in.
Exercise:
Problem:
17 mi
Solution:
367.5 sq mi
Exercise:
Problem:
(approximate)
Solution:
452.16 sq ft
Exercise:
Problem:
—
2cm 5 cm
Solution:
44.28 sq cm
The Meaning and Notation for Volume
The product (length unit)(length unit) (length unit) = (length unit)°, or
cubic length unit (cu length unit), can be interpreted physically as the volume of
a three-dimensional object.
Volume
The volume of an object is the amount of cubic length units contained in the
object.
For example, 4 cu mm means that 4 cubes, 1 mm on each side, would precisely
fill some three-dimensional object. (The cubes may have to be cut and
rearranged so they match the shape of the object.)
r
h = height
; g
w = width
——— /=lengthh —>
Rectangular solid
Sphere
h = height
Cylinder
|
:
h = hdight
/ | \
J |
|
( |
-_—______—>}
r= radius a
ie re ed te
Cone
Volume Formulas
Figure Volume Formula
Rectangular Ve = l-weh
solid = (area of base) - (height)
Sphere Vo= - ae ae
Statement
The
volume of
a
rectangular
solid is the
length
times the
width
times the
height.
The
volume of
Cvlind Voy) = mer? eh
yunder = (area of base) - (height)
V.. — 7" “WT: r2 . h
Cone
= (area of base) - (height)
Finding Volumes of Some Common Geometric Objects
Sample Set B
Example:
Find the volume of the rectangular solid.
a sphere is
$ times 7
times the
cube of the
radius.
The
volume of
a cylinder
is 7 times
the square
of the
radius
times the
height.
The
volume of
a cone is
$ times 7
times the
square of
the radius
times the
height.
3 in.
10 in.
9 in.
Ve = baa
= Q9in.-10in.- 3 in.
= 2; Uicumin:
= 270in.®
The volume of this rectangular solid is 270 cu in.
Example:
Find the approximate volume of the sphere.
Vg = ~ Pas oF
= (4) - (3.14) -(6 cm)’
~ (4) - (3.14) - (216 cu cm)
~ 904.32 cucm
The approximate volume of this sphere is 904.32 cu cm, which is often written
as 904.32 cm’.
Example:
Find the approximate volume of the cylinder.
ng
Voy
2
2
NN
law)
4.9 ft
7.8 ft
n-r2-h
(3149 fe) (7.8 tt)
(3.14) - (24.01 sq ft) - (7.8 ft)
(3.14) - (187.278 cu ft)
588.05292 cu ft
The volume of this cylinder is approximately 588.05292 cu ft. The volume is
approximate because we approximated 7 with 3.14.
Example:
Find the approximate volume of the cone. Round to two decimal places.
Ve. = aa r*-h
(3.14) - (2mm)? - (5 mm)
+) - (3.14) - (4.sq mm) - (5 mm)
) - (3.14) - (20 cu mm)
= |
G
~ 20.93 cumm
=~ 20.93 cumm
The volume of this cone is approximately 20.93 cu mm. The volume is
approximate because we approximated 7 with 3.14.
Practice Set B
Find the volume of each geometric object. If a is required, approximate it with
3.14 and find the approximate volume.
Exercise:
Problem:
3 in
f
— 10 in.
9 in.
Solution:
21 cu in.
Exercise:
Problem: Sphere
Solution:
904.32 cu ft
Exercise:
Problem:
Solution:
157 cum
Exercise:
Problem:
Solution:
0.00942 cu in.
Exercises
Find each indicated measurement.
Exercise:
Problem: Area
2m
8 m
Solution:
16 sqm
Exercise:
Problem: Area
2.3 in.
4.1 in.
Exercise:
Problem: Area
1.1 mm
Solution:
1.21 sq mm
Exercise:
Problem: Area
8 cm
Exercise:
Problem: Area
4 in.
Solution:
18 sq in.
Exercise:
Problem: Area
20 cm
Exercise:
Problem: Exact area
22 ft
Solution:
(60.57 + 132) sq ft
Exercise:
Problem: Approximate area
6 ft
=
Exercise:
Problem: Area
10.2 in.
Solution:
40.8 sq in.
Exercise:
Problem: Area
15 mm
Exercise:
Problem: Approximate area
8.4 in.
Solution:
31.0132 sq in.
Exercise:
Problem: Exact area
Exercise:
Problem: Approximate area
Solution:
158.2874 sq mm
Exercise:
Problem: Exact area
19 cm
Exercise:
Problem: Approximate area
| : :
(3.2 in. 9.4 in.
| |
6.1 In.
Solution:
64.2668 sq in.
Exercise:
Problem: Area
1.74 in.
5.21 in.
Exercise:
Problem: Approximate area
Solution:
43.96 sq ft
Exercise:
Problem: Volume
Exercise:
Problem: Volume
8 mm
Solution:
512 cucm
Exercise:
Problem: Exact volume
PE = Se
ge
4
/ 3 i fr
o> in.
/ 4 \
| me i sphere
\ /
Exercise:
Problem: Approximate volume
_14em
— sphere
/
j
\ /
Mia alt
Solution:
11.49 cu cm
Exercise:
Problem: Approximate volume
0.9 ft
Exercise:
Problem: Exact volume
Solution:
1A cu ft
Exercise:
Problem: Approximate volume
9.2 in.
Exercise:
Problem: Approximate volume
Solution:
22.08 cu in.
Exercise:
Problem: Approximate volume
8.1 fi
Exercises for Review
Exercise:
Problem: ({link]) In the number 23,426, how many hundreds are there?
Solution:
4
Exercise:
Problem: ({link]) List all the factors of 32.
Exercise:
Problem: ({link]) Find the value of 42 — 32. + 12.
Solution:
31 5 7 _
31 __ 97 — 2.58
Exercise:
i
Problem: ({link]) Find the value of ots
242°
Exercise:
Problem: ({link]) Find the perimeter.
ae
os ™
1.2 m/ 12.4 m
e4 a
Fa Sc
: _ OS
8.3 m
Solution:
27.9m
Summary of Key Concepts
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module reviews the key concepts from the chapter
"Measurement and Geometry."
Summary of Key Concepts
Measurement ((link])
Measurement is comparison to some standard.
Standard Unit of Measure ([link])
A quantity that is used for comparison is called a standard unit of
measure.
Two Types of Measurement Systems ([link])
There are two major types of measurement systems in use today. They are
the United States system and the metric system.
Unit Fraction ({link])
A unit fraction is a fraction that has a value of 1. Unit fractions can be used
to convert from one unit of measure to another.
Meter, Liter, Gram, and associated prefixes ({link])
Common units of measure in the metric system are the meter (m), for
length, the liter (L), for volume, and the gram (g), for mass. To each of
these units, a prefix can be attached.
e kilothousand
e decitenth
e hectohundred
e centihundredth
e dekaten
e millithousandth
Metric Conversions ({link])
To convert from one metric unit to another:
1. Determine the location of the original number on the metric scale.
2. Move the decimal point of the original number in the same direction
and the same number of places as is necessary to move to the metric
unit you wish to convert to.
Denominate Numbers ([link])
Numbers that have units of measure associated with them are denominate
numbers. The number 25 mg is a denominate number since the mg unit is
associated with the pure number 25. The number 82 is not a denominate
number since it has no unit of measure associated with it.
Simplified Denominate Number ((link])
A denominate number is simplified when the number of standard units of
measure associated with it does not exceed the next higher type of unit. 55
min is simplified, whereas 65 min is not simplified
Addition and Subtraction of Denominate Numbers ((link])
Denominate numbers can be added or subtracted by
1. writing the numbers vertically so that the like units appear in the same
column.
2. adding or subtracting the number parts, carrying along the unit.
3. simplifying the sum or difference.
Multiplying a Denominate Number by a Whole Number ([link])
To multiply a denominate number by a whole number, multiply the number
part of each unit by the whole number and affix the unit to the product.
Dividing a Denominate Number by a Whole Number ((link])
To divide a denominate number by a whole number, divide the number part
of each unit by the whole number beginning with the largest unit. Affix the
unit to this quotient. Carry the remainder to the next unit.
Polygon ([link])
A polygon is a closed plane (flat) figure whose sides are line segments
(portions of straight lines).
Perimeter ([{link])
The perimeter of a polygon is the distance around the polygon.
Circumference, Diameter, Radius ({link])
The circumference of a circle is the distance around the circle. The
diameter of a circle is any line segment that passes through the center of
the circle and has its endpoints on the circle. The radius of a circle is one
half the diameter of the circle.
The number z ([{link])
The symbol 7, read "pi," represents the nonterminating, nonrepeating
decimal number 3.14159... . For computational purposes, 7 is often
approximated by the number 3.14.
Formula ({link])
A formula is a rule for performing a task. In mathematics, a formula is a
rule that directs us in computations.
Circumference Formulas ({link])
C=7-dC x (3.14)d
C=2¢a-¢ Ce 2(3.14)r
Area ([link])
The area of a surface is the amount of square length units contained in the
surface.
Volume ([link])
The volume of an object is a measure of the amount of cubic length units
contained in the object.
Area Formulas ([link])
Triangle: A = ; -b-h
Rectangle: A = /- w
Parallelogram: A = b-h
Trapezoid: A = + - (b; + bz) +h
Circle: A= 7-r?
Volume Formulas ([link])
Rectangle solid: V =/-w-h
Sphere: V = - 77?
Cylinder: V = 7-
r2-h
Cone: V = 3 -7-1r?-h
Exercise Supplement
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is an exercise supplement for the chapter
"Measurement and Geometry" and contains many exercise problems. Odd
problems are accompanied by solutions.
Exercise Supplement
Measurement and the United States System ((link])
Exercise:
Problem: What is measurement?
Solution:
Measurement is comparison to a standard (unit of measure).
For problems 2-6, make each conversion. Use the conversion table given in
section 9.1.
Exercise:
Problem: 9 ft= yd
Exercise:
Problem: 32 oz= |b
Solution:
2 pounds
Exercise:
Problem: 1,500 mg = g
Exercise:
Problem: 12,000 lb = T
Solution:
6 tons
Exercise:
Problem: 5,280 ft= mi
For problems 7-23, make each conversion.
Exercise:
Problem: 23 yd to ft
Solution:
69 feet
Exercise:
Problem: 25mi to yd
Exercise:
Problem: 8 in. to ft
Solution:
+ = 0.666 feet
Exercise:
Problem: 51 in. to mi
Exercise:
Problem: 3 gt to pt
Solution:
6 pints
Exercise:
Problem: 8 |b to oz
Exercise:
Problem: 5 cups to tbsp
Solution:
80 tablespoons
Exercise:
Problem: 9 da to hr
Exercise:
Problem: 35 min to sec
Solution:
210 seconds
Exercise:
Problem: 3. wk to min
The Metric System of Measurement ({link])
Exercise:
Problem: 250 mL to L
Solution:
1 _
+ = 0.25L
Exercise:
Problem: 18.57 cm to m
Exercise:
Problem: 0.01961 kg to mg
Solution:
19,610 mg
Exercise:
Problem: 52,211 mg to kg
Exercise:
Problem: 54.006 dag to g
Solution:
540.06 g
Exercise:
Problem: 1.181 hg to mg
Exercise:
Problem: 3.5 kL to mL
Solution:
3,900,000 mL
Simplification of Denominate Numbers ((link])
For problems 24-31, perform the indicated operations. Simplify, if possible.
Exercise:
Problem: Add 8 min 50 sec to 5 min 25 sec.
Exercise:
Problem: Add 3 wk 3 da to 2 wk 5 da
Solution:
6 weeks 1 day
Exercise:
Problem: Subtract 4 gal 3 gt from 5 gal 2 at.
Exercise:
Problem: Subtract 2 gal 3 gt 1pt from 8 gal 2 qt.
Solution:
5 gallons 2 quarts 1 pint
Exercise:
Problem: Subtract 5 wk 4 da 21 hr from 12 wk 3 da 14 hr.
Exercise:
Problem: Subtract 2 T 1,850 lb from 10 T 1,700 lb.
Solution:
7 T 1,850 pounds
Exercise:
Problem:
Subtract the sum of 2 wk 3 da 15 hr and 5 wk 2 da 9 hr from 10 wk.
Exercise:
Problem:
Subtract the sum of 20 hr 15 min and 18 hr 18 min from the sum of 8
da 1 hr 16 min 5 sec.
Solution:
7 days, 11 hours, 56 minutes, 7 seconds
For problems 32-43, simplify, if necessary.
Exercise:
Problem: 18 in.
Exercise:
Problem: 4 ft
Solution:
1 yard 1 foot
Exercise:
Problem: 23 da
Exercise:
Problem: 3,100 |b
Solution:
1 ton 1,100 pounds
Exercise:
Problem: 135 min
Exercise:
Problem: 4 tsp
Solution:
1 tablespoon 1 teaspoon
Exercise:
Problem: 10 fl oz
Exercise:
Problem: 7 pt
Solution:
3 quarts 1 pint
Exercise:
Problem: 9 gt
Exercise:
Problem: 2,300 mm
Solution:
2.3 meters
Exercise:
Problem: 14,780 mL
Exercise:
Problem: 1,050 m
Solution:
1.05 km
Perimeter, Circumference, Area and Volume of Geometric Figures and
Objects ([link],[link])
For problems 44-58, find the perimeter, circumference, area or volume.
Exercise:
Problem: Perimeter, area
20 mm
50 mm
Exercise:
Problem: Approximate circumference
Solution:
9.652 sq cm
Exercise:
Problem: Approximate volume
Sees
=~
>>
5
me
a
\
\
|
|
| let
os
wl
2
M
Exercise:
Problem: Approximate volume
a ot) 4
Solution:
104.28568 cu ft
Exercise:
Problem: Exact area
Exercise:
Problem: Exact area
0.6 in.
Solution:
0.187 sq in.
Exercise:
Problem: Exact volume
a
4 \
_
aie:
(10.1 cm
iar 5S
: “4.2 cm
Exercise:
Problem: Approximate volume
\
ae To \
—
\
— __| sphere
Solution:
267.94667 cu mm
Exercise:
Problem: Area
4.1 in.
| \
3.9 in.
5.1 in.
Exercise:
Problem: Volume
2 cm
2 cm
8 cm
Solution:
32 cucm
Exercise:
Problem: Exact area
J
fo \
a |
“ —,
ye . |
/ oa
~N
6 mm
Exercise:
Problem: Approximate area
12 in.
Solution:
39.48 sq in.
Exercise:
Problem: Exact area
Exercise:
Problem: Approximate area
ad Sa
ft ae.
/ ( a
yf / Bi,
f ¥ 4
ff
S
4
Solution:
56.52 sq ft
Exercise:
Problem: Approximate area
Proficiency Exam
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is a proficiency exam to the chapter
"Measurement and Geometry." Each problem is accompanied with a
reference link pointing back to the module that discusses the type of
problem demonstrated in the question. The problems in this exam are
accompanied by solutions.
Proficiency Exam
Exercise:
Problem:
({link]) The process of determining, by comparison to some standard,
the size of something is called .
Solution:
measurement
For problems 2-9, make each conversion.
Exercise:
Problem: ({link]) 14 yards to feet
Solution:
A? feet
Exercise:
Problem: ([link]) 51 feet to inches
Solution:
612 inches
Exercise:
Problem: ({link]) z yard to feet
Solution:
1 foot
Exercise:
Problem: ({link}) 24 minutes to seconds
Solution:
135 seconds
Exercise:
Problem: ([link]) 8,500 mg to cg
Solution:
850 cg
Exercise:
Problem: ((link]) 5.8623 L to kL
Solution:
0.0058623 kL
Exercise:
Problem: ({link]) 213.1062 mm to m
Solution:
0.2132062 m
Exercise:
Problem: ({link]) 100,001 kL to mL
Solution:
100,001,000,000 mL
For problems 10-13, simplify each number.
Exercise:
Problem: ({link]) 23 da
Solution:
3 weeks 2 days
Exercise:
Problem: ({link]) 88 ft
Solution:
29 yards 1 foot
Exercise:
Problem: ({link]) 4216 lb
Solution:
2 tons 216 pounds
Exercise:
Problem: ({link]) 7 gt
Solution:
1 gallon 3 quarts
For problems 14-18, perform the indicated operations. Simplify answers if
possible.
Exercise:
Problem: ([link]) Add 6 wk 3 da to 2 wk 2 da.
Solution:
8 weeks 5 days
Exercise:
Problem: ([link]) Add 9 gal 3 qt to 4 gal 3 at.
Solution:
14 gallons 2 quarts
Exercise:
Problem: ([link]) Subtract 3 yd 2 ft 5 in. from 5 yd 8 ft 2 in.
Solution:
2 yards 5 feet 9 inches
Exercise:
Problem: ({link]) Subtract 2 hr 50 min 12 sec from 3 hr 20 min 8 sec.
Solution:
29 minutes 56 seconds
Exercise:
Problem:
({link]) Subtract the sum of 3 wk 6 da and 2 wk 3 da from 10 wk.
Solution:
3 weeks 5 days
For problems 19-30, find either the perimeter, circumference, area, or
volume.
Exercise:
Problem: ({link]) Perimeter
8.61 m
8.61 m
Solution:
34.44 m
Exercise:
Problem: ({link]) Perimeter
7.2 mm
5.8 mm / Ng 9 mm
/ Sy,
{ my
15.1 mm
Solution:
36 mm
Exercise:
Problem: ({link]) Approximate circumference
> oe =
f \
4 \,
/ 14 ft \
/ an
fe
\
Sig
Solution:
87.92 feet
Exercise:
Problem: ({link]) Approximate perimeter
5 mi Lmi 45 mi
15 mi
12 mi
Solution:
55.14 miles
Exercise:
Problem: ({link]) Area
1.5 in.
Solution:
3.75 sq in.
Exercise:
Problem: ({link]) Approximate area
Solution:
6.002826 sq cm
Exercise:
Problem: ({link]) Approximate area
Solution:
6.28 sq miles
Exercise:
Problem: ({link]) Area
4 in.
9 in.
Solution:
13 sq in.
Exercise:
Problem: ({link]) Exact area
Solution:
84.64 sq in.
Exercise:
Problem: ({link]) Approximate volume
Solution:
25.12 cumm
Exercise:
Problem: ({link]) Exact volume
Solution:
4.608 cu ft
Exercise:
Problem: ({link]) Approximate volume
2 mm
Solution:
340.48 cu mm
Objectives
This module contains the learning objectives for the chapter "Signed
Numbers" from Fundamentals of Mathematics by Denny Burzynski and
Wade Ellis, jr.
After completing this chapter, you should
Variables, Constants, and Real Numbers ([{link])
e be able to distinguish between variables and constants
e be able to recognize a real number and particular subsets of the real
numbers
e understand the ordering of the real numbers
Signed Numbers ((link])
e be able to distinguish between positive and negative real numbers
e be able to read signed numbers
e understand the origin and use of the double-negative product property
Absolute Value ({link]})
¢ understand the geometric and algebraic definitions of absolute value
Addition of Signed Numbers ((link])
e be able to add numbers with like signs and with unlike signs
e be able to use the calculator for addition of signed numbers
Subtraction of Signed Numbers (({link])
e understand the definition of subtraction
e be able to subtract signed numbers
e be able to use a calculator to subtract signed numbers
Multiplication and Division of Signed Numbers ({link])
e be able to multiply and divide signed numbers
e be able to multiply and divide signed numbers using a calculator
Variables, Constants, and Real Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses variables, constants, and real
numbers. By the end of the module students should be able to distinguish
between variables and constants, be able to recognize a real number and
particular subsets of the real numbers and understand the ordering of the
real numbers.
Section Overview
e Variables and Constants
e Real Numbers
e Subsets of Real Numbers
e Ordering Real Numbers
Variables and Constants
A basic distinction between algebra and arithmetic is the use of symbols
(usually letters) in algebra to represent numbers. So, algebra is a
generalization of arithmetic. Let us look at two examples of situations in
which letters are substituted for numbers:
1. Suppose that a student is taking four college classes, and each class
can have at most 1 exam per week. In any 1-week period, the student
may have 0, 1, 2, 3, or 4 exams. In algebra, we can let the letter x
represent the number of exams this student may have in a 1-week
period. The letter x may assume any of the various values 0, 1, 2, 3, 4.
2. Suppose that in writing a term paper for a biology class a student needs
to specify the average lifetime, in days, of a male housefly. If she does
not know this number off the top of her head, she might represent it (at
least temporarily) on her paper with the letter t (which reminds her of
time). Later, she could look up the average time in a reference book
and find it to be 17 days. The letter ¢ can assume only the one value,
17, and no other values. The value t is constant.
Variable, Constant
1. A letter or symbol that represents any member of a collection of two or
more numbers is called a variable.
2. A letter or symbol that represents one specific number, known or
unknown, is called a constant.
In example 1, the letter zx is a variable since it can represent any of the
numbers 0, 1, 2, 3, 4. The letter t example 2 is a constant since it can only
have the value 17.
Real Numbers
Real Number Line
The study of mathematics requires the use of several collections of
numbers. The real number line allows us to visually display (graph) the
numbers in which we are interested.
A line is composed of infinitely many points. To each point we can
associate a unique number, and with each number, we can associate a
particular point.
Coordinate
The number associated with a point on the number line is called the
coordinate of the point.
Graph
The point on a number line that is associated with a particular number is
called the graph of that number.
Constructing a Real Number Line
We construct a real number line as follows:
1. Draw a horizontal line.
2. Origin
Choose any point on the line and label it 0. This point is called the
origin.
see
0
3. Choose a convenient length. Starting at 0, mark this length off in both
directions, being careful to have the lengths look like they are about
the same.
+++ ++ +--+ _ + + _ ++
0
We now define a real number.
Real Number
A real number is any number that is the coordinate of a point on the real
number line.
Positive Numbers, Negative Numbers
Real numbers whose graphs are to the right of 0 are called positive real
numbers, or more simply, positive numbers. Real numbers whose graphs
appear to the left of 0 are called negative real numbers, or more simply,
negative numbers.
$$$ "=
0 ux~ YY
Negative Positive
numbers numbers
The number 0 is neither positive nor negative.
Subsets of Real Numbers
The set of real numbers has many subsets. Some of the subsets that are of
interest in the study of algebra are listed below along with their notations
and graphs.
Natural Numbers, Counting Numbers
The natural or counting numbers (JV): 1, 2, 3, 4, ... Read “and so on.”
Whole Numbers
The whole numbers (W): 0, 1, 2, 3, 4,...
Notice that every natural number is a whole number.
Integers
The integers (Z): ... -3, -2, -1, 0, 1, 2, 3,...
+—+—+ 6444644 >
-4 -3-2-1 012 8 4
Notice that every whole number is an integer.
Rational Numbers (Fractions)
The rational numbers (Q): Rational numbers are sometimes called
fractions. They are numbers that can be written as the quotient of two
integers. They have decimal representations that either terminate or do not
terminate but contain a repeating block of digits. Some examples are
=3 qt es
ZF = 0.75 85 = 8.407407407 ...
Terminating Nonterminating, but repeating
Some rational numbers are graphed below.
-3-2-1 012 8 4
ee
— 3% —-3 € 2=# 33
Notice that every integer is a rational number.
Notice that there are still a great many points on the number line that have
not yet been assigned a type of number. We will not examine these other
types of numbers in this text. They are examined in detail in algebra. An
example of these numbers is the number 7, whose decimal representation
does not terminate nor contain a repeating block of digits. An
approximation for 7 is 3.14.
Sample Set A
Example:
Is every whole number a natural number?
No. The number 0 is a whole number but it is not a natural number.
Example:
Is there an integer that is not a natural number?
Yes. Some examples are 0, -1, -2, -3, and -4.
Example:
Is there an integer that is a whole number?
Yes. In fact, every whole number is an integer.
Practice Set A
Exercise:
Problem: Is every natural number a whole number?
Solution:
yes
Exercise:
Problem
: Is every whole number an integer?
Solution:
yes
Exercise:
Problem
: Is every integer a real number?
Solution:
yes
Exercise:
Problem
: Is there an integer that is a whole number?
Solution:
yes
Exercise:
Problem
: Is there an integer that is not a natural number?
Solution:
yes
Ordering
Real Numbers
Ordering Real Numbers
A real number 0 is said to be greater than a real number a, denoted b > a,
if b is to the right of a on the number line. Thus, as we would expect, 5 > 2
since 5 is to the right of 2 on the number line. Also, —2 > —5 since -2 is to
the right of -5 on the number line.
-2>-5 5 >2
-§ -4 -3 -2 -1 0 1 2 3 4 6&5
If we let a and b represent two numbers, then a and 6 are related in exactly
one of three ways: Either
Equality Symbol
a=b aandbareequal (8 = 8)
Inequality Symbols
a>b a is greater than b (8 > 5)
a<b a isless than b (5 < 8)
Some variations of these symbols are
a#b a isnot equal to b (8 #5)
a>bd a is greater than or equal to b (a > 8)
a<b a isless than or equal to b (a < 8)
Sample Set B
Example:
What integers can replace x so that the following statement is true?
=o Se 2
-§ -4 -3 -2-1 0 1 2 83 4 65
The integers are -3, -2, -1, 0, 1.
Example:
Draw a number line that extends from -3 to 5. Place points at all whole
numbers between and including -1 and 3.
-S:e2-=1 -@ 1 2 3.0: 5
-1 is not a whole number
Practice Set B
Exercise:
Problem:
What integers can replace x so that the following statement is true?
—h5<2<2
Solution:
-5, -4, -3, -2, -1, 0
Exercise:
Problem:
Draw a number line that extends from -4 to 3. Place points at all
natural numbers between, but not including, -2 to 2.
Solution:
-4-3-2-1 01 2 8
Exercises
For the following 8problems, next to each real number, note all collections
to which it belongs by writing NV for natural number, W for whole number,
or Z for integer. Some numbers may belong to more than one collection.
Exercise:
Problem: 6
Solution:
N,W,Z
Exercise:
Problem: 12
Exercise:
Problem: 0
Solution:
W,Z
Exercise:
Problem: 1
Exercise:
Problem: -3
Solution:
Z
Exercise:
Problem: -7
Exercise:
Problem: -805
Solution:
Z
Exercise:
Problem: -900
Exercise:
Problem:
Is the number 0 a positive number, a negative number, neither, or both?
Solution:
Neither
Exercise:
Problem:
An integer is an even integer if it is evenly divisible by 2. Draw a
number line that extends from -5 to 5 and place points at all negative
even integers and all positive odd integers.
Exercise:
Problem:
Draw a number line that extends from -5 to 5. Place points at all
integers that satisfy —3 < xa < 4.
Solution:
-5 -4 -3 -2 -1 0 1 2 83 4 65
Exercise:
Problem: Is there a largest two digit number? If so, what is it?
Exercise:
Problem: Is there a smallest two digit number? If so, what is it?
Solution:
Yes, 10
For the pairs of real numbers in the following 5 problems, write the
appropriate symbol (<, >, =) in place of the 0.
Exercise:
Problem: -7 1 -2
Exercise:
Problem: -5 10
Solution:
<
Exercise:
Problem: -1 14
Exercise:
Problem: 6 0 -1
Solution:
>
Exercise:
Problem: 10 0 10
For the following 5 problems, what numbers can replace m so that the
following statements are true?
Exercise:
Problem: —1 < m < —5, m an integer.
Solution:
{-1, 0, il Zs 3; 4, oO}
Exercise:
Problem: —7 < m < —1, man integer.
Exercise:
Problem: —3 < m < 2, ma natural number.
Solution:
{1}
Exercise:
Problem: —15 < m < —1,™m™ a natural number.
Exercise:
Problem: —5 < m < 5, ma whole number.
Solution:
{0, 1, 2, 3, 4}
For the following 10 problems, on the number line, how many units are
there between the given pair of numbers?
Exercise:
Problem
Exercise:
Problem:
: 0 and 3
-4 and 0
Solution:
4
Exercise:
Problem:
Exercise:
Problem:
-1 and6
-6 and 2
Solution:
8
Exercise:
Problem:
Exercise:
Problem:
-3 and 3
Are all positive numbers greater than zero?
Solution:
yes
Exercise:
Problem
Exercise:
Problem
: Are all positive numbers greater than all negative numbers?
: Is O greater than all negative number?
Solution:
yes
Exercise:
Problem
Exercise:
Problem
: Is there a largest natural number?
: Is there a largest negative integer?
Solution:
yes, -1
Exercises for Review
Exercise:
Problem: ({link]) Convert 62 to an improper fraction.
Exercise:
Problem: ([link]) Find the value: 3 of 33
Solution:
9 4
— Or 1 For 1.8
Exercise:
Problem: ({link]) Find the sum of $ + 2.
Exercise:
Problem: ({link]) Convert 30.06 cm to m.
Solution:
0.3006 m
Exercise:
Problem: ({link]) Find the area of the triangle.
3 mm[
16 mm
Signed Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade
Ellis, Jr. This module discusses signed numbers. By the end of the module students be
able to distinguish between positive and negative real numbers, be able to read signed
numbers and understand the origin and use of the double-negative product property.
Section Overview
e Positive and Negative Numbers
e Reading Signed Numbers
¢ Opposites
¢ The Double-Negative Property
Positive and Negative Numbers
Positive and Negative Numbers
Each real number other than zero has a sign associated with it. A real number is said
to be a positive number if it is to the right of 0 on the number line and negative if it
is to the left of 0 on the number line.
Note:
THE NOTATION OF SIGNED NUMBERS
+ and — Notation
A number is denoted as positive if it is directly preceded by a plus sign or no sign at
all.
A number is denoted as negative if it is directly preceded by a minus sign.
Reading Signed Numbers
The plus and minus signs now have two meanings:
The plus sign can denote the operation of addition or a positive number.
The minus sign can denote the operation of subtraction or a negative number.
To avoid any confusion between "sign" and "operation," it is preferable to read the
sign of a number as "positive" or "negative." When "+" is used as an operation sign, it
is read as "plus." When "—" is used as an operation sign, it is read as "minus."
Sample Set A
Read each expression so as to avoid confusion between "operation" and "sign."
Example:
—8 should be read as "negative eight" rather than "minus eight."
Example:
—6 + (—3)should be read as "negative six plus negative three" rather than "minus
six plus minus three."
Example:
—15 — (—6)should be read as "negative fifteen minus negative six" rather than
"minus fifteen minus minus six."
Example:
—5-+ 7 should be read as "negative five plus seven" rather than "minus five plus
seven."
Example:
0 — 2 should be read as "zero minus two."
Practice Set A
Write each expression in words.
Exercise:
Problem: 6 + 1
Solution:
six plus one
Exercise:
Problem: 2 + (—8)
Solution:
two plus negative eight
Exercise:
Problem: —7 + 5
Solution:
negative seven plus five
Exercise:
Problem: —10 — (+3)
Solution:
negative ten minus three
Exercise:
Problem: —1 — (—8)
Solution:
negative one minus negative eight
Exercise:
Problem: 0 + (—11)
Solution:
zero plus negative eleven
Opposites
Opposites
On the number line, each real number, other than zero, has an image on the opposite
side of 0. For this reason, we say that each real number has an opposite. Opposites
are the same distance from zero but have opposite signs.
The opposite of a real number is denoted by placing a negative sign directly in front
of the number. Thus, if @ is any real number, then —a is its opposite.
Note:The letter "a" is a variable. Thus, "a" need not be positive, and "—a" need not
be negative.
If a is any real number, —a is opposite a on the number line.
a positive a negative
——+—+—, +—-+—1+—-t+
-a 0 a a O -a
The Double-Negative Property
The number a is opposite —a on the number line. Therefore, —(—a) is opposite —a
on the number line. This means that
—(—a)=a
From this property of opposites, we can suggest the double-negative property for real
numbers.
Double-Negative Property: —(—a) =a
If a is a real number, then
—(-a)=a
Sample Set B
Find the opposite of each number.
Example:
If a = 2, then —a = —2. Also, —(—a) = —(—2) = 2.
-2 0 2
—a a
—(-a)
Example:
If a = —4, then —a = —(—4) = 4. Also, —(—a) =a =
——_—_+—
-4 0 4
a —a
Practice Set B
Find the opposite of each number.
Exercise:
Problem: 8
Solution:
-8
Exercise:
Problem: 17
Solution:
-17
Exercise:
Problem:
Solution:
6
Exercise:
Problem:
Solution:
15
Exercise:
Problem:
Solution:
-1
Exercise:
Problem:
Solution:
7
Exercise:
Problem:
Solution:
-15
Suppose a is a positive number. Is —a positive or negative?
—a is negative
Exercise:
Problem: Suppose a is a negative number. Is —a positive or negative?
Solution:
—a is positive
Exercise:
Problem:
Suppose we do not know the sign of the number k. Is —k positive, negative, or
do we not know?
Solution:
We must say that we do not know.
Exercises
Exercise:
Problem: A number is denoted as positive if it is directly preceded by .
Solution:
+ (or no sign)
Exercise:
Problem: A number is denoted as negative if it is directly preceded by .
How should the number in the following 6 problems be read? (Write in words.)
Exercise:
Problem: —7
Solution:
negative seven
Exercise:
Problem: —5
Exercise:
Problem: 15
Solution:
fifteen
Exercise:
Problem: 11
Exercise:
Problem: —(—1)
Solution:
negative negative one, or opposite negative one
Exercise:
Problem: — (—5)
For the following 6 problems, write each expression in words.
Exercise:
Problem: 5 + 3
Solution:
five plus three
Exercise:
Problem: 3 + 8
Exercise:
Problem: 15 + (—3)
Solution:
fifteen plus negative three
Exercise:
Problem: 1 + (—9)
Exercise:
Problem: —7 — (—2)
Solution:
negative seven minus negative two
Exercise:
Problem: 0 — (—12)
For the following 6 problems, rewrite each number in simpler form.
Exercise:
Problem: —(—2)
Solution:
2
Exercise:
Problem: —(—16)
Exercise:
Problem: —|—(—8)|
Solution:
-8
Exercise:
Problem: — |—(—20)|
Exercise:
Problem: 7 — (—3)
Solution:
7+3=10
Exercise:
Problem: 6 — (—4)
Exercises for Review
Exercise:
Problem: ({link]) Find the quotient; 8+27.
Solution:
0.296
Exercise:
Problem: (({link]) Solve the proportion: ~ =>
Exercise:
60
Problem: (({link]) Use the method of rounding to estimate the sum: 5829 + 8767
Solution:
6,000 + 9,000 = 15 000 (5,829 + 8,767 = 14.596) or 5,800 + 8,800 = 14 600
Exercise:
Problem: ((link]) Use a unit fraction to convert 4 yd to feet.
Exercise:
Problem: ((link]) Convert 25 cm to hm.
Solution:
0.0025 hm
Absolute Value
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses absolute value. By the end of the
module students should understand the geometric and algebraic definitions
of absolute value.
Section Overview
e Geometric Definition of Absolute Value
e Algebraic Definition of Absolute Value
Geometric Definition of Absolute Value
Absolute Value-Geometric Approach
Geometric definition of absolute value:
The absolute value of a number a, denoted | a
on the number line.
, is the distance from a to 0
Absolute value answers the question of "how far," and not "which way."
The phrase "how far" implies "length" and length is always a nonnegative
quantity. Thus, the absolute value of a number is a nonnegative number.
Sample Set A
Determine each value.
Example:
[eae
4 units in length
Example:
|-4|=4
4 units in length
SSS
-~6 -5 -4 -3 -2 -1 0
Example:
HOE
Example:
— | 5 |= —5. The quantity on the left side of the equal sign is read as
"negative the absolute value of 5." The absolute value of 5 is 5. Hence,
negative the absolute value of 5 is -5.
Example:
— | —3 |= —3. The quantity on the left side of the equal sign is read as
"negative the absolute value of -3." The absolute value of -3 is 3. Hence,
negative the absolute value of -3 is —(3) = —3.
Practice Set A
By reasoning geometrically, determine each absolute value.
Exercise:
Problem: | 7 |
Solution:
x
Exercise:
Problem
:| —3 |
Solution:
3
Exercise:
Problem
a |
Solution:
12
Exercise:
Problem:
0 |
Solution:
0
Exercise:
Problem
:— |9|
Solution:
-9
Exercise:
Problem
:— | -6 |
Solution:
-6
Algebraic Definition of Absolute Value
From the problems in [link], we can suggest the following algebraic defini-
tion of absolute value. Note that the definition has two parts.
Absolute Value—Algebraic Approach
Algebraic definition of absolute value
The absolute value of a number a is
a, ifa>0O
ja| = .
—a, if <0
The algebraic definition takes into account the fact that the number a could
be either positive or zero (a > 0) or negative (a < 0).
1. If the number a is positive or zero (a > 0), the upper part of the
definition applies. The upper part of the definition tells us that if the
number enclosed in the absolute value bars is a nonnegative number,
the absolute value of the number is the number itself.
2. The lower part of the definition tells us that if the number enclosed
within the absolute value bars is a negative number, the absolute value
of the number is the opposite of the number. The opposite of a
negative number is a positive number.
Note:The definition says that the vertical absolute value lines may be
eliminated only if we know whether the number inside is positive or
negative.
Sample Set B
Use the algebraic definition of absolute value to find the following values.
Example:
| 8 |. The number enclosed within the absolute value bars is a nonnegative
number, so the upper part of the definition applies. This part says that the
absolute value of 8 is 8 itself.
|8|=8
Example:
| —3 |. The number enclosed within absolute value bars is a negative
number, so the lower part of the definition applies. This part says that the
absolute value of -3 is the opposite of -3, which is —(—3). By the
definition of absolute value and the double-negative property,
lease ect re eee
Practice Set B
Use the algebraic definition of absolute value to find the following values.
Exercise:
Problem: | 7 |
Solution:
7
Exercise:
Problem: | 9 |
Solution:
a
Exercise:
Problem: | —12 |
Solution:
12
Exercise:
Problem: | —5 |
Solution:
3)
Exercise:
Problem: — | 8 |
Solution:
-8
Exercise:
Problem: — | 1 |
Solution:
=i
Exercise:
Problem: — | —52 |
Solution:
-52
Exercise:
Problem: — | —31 |
Solution:
-31
Exercises
Determine each of the values.
Exercise:
Problem: | 5 |
Solution:
5
Exercise:
Problem: | 3 |
Exercise:
Problem: | 6 |
Solution:
6
Exercise:
Problem: | —9 |
Exercise:
Problem:
fe
Solution:
1
Exercise:
Problem
Exercise:
Problem:
| —4|
—|3|
Solution:
-3
Exercise:
Problem
Exercise:
Problem
:—|7|
:— |-14|
Solution:
-14
Exercise:
Problem:
Exercise:
Problem
0 |
>| —26 |
Solution:
26
Exercise:
Problem: — | —26 |
Exercise:
Problem: —(— | 4 |)
Solution:
4
Exercise:
Problem: —(— | 2 |)
Exercise:
Problem: —(— | —6 |)
Solution:
6
Exercise:
Problem: —(— | —42 |)
Exercise:
Problem: | 5 | — | —2 |
Solution:
3
Exercise:
Problem: | —2 h
Exercise:
Problem: | —(2 - 3) |
Solution:
6
Exercise:
Problem: | —2 | — | —9 |
Exercise:
Problem: (| —6 | + | 4 1)”
Solution:
100
Exercise:
Problem: (| —1 | — | 1 \)°
Exercise:
2 3
Problem: (| 4 | + | —6 |)" — (| —2 |)
Solution:
2
Exercise:
Problem: —{|—10| — 6)”
Exercise:
gre
Problem: -{-[- [=4|] + | -3 |] \
Solution:
-1
Exercise:
Problem:
A Mission Control Officer at Cape Canaveral makes the statement
“lift-off, T minus 50 seconds.” How long is it before lift-off?
Exercise:
Problem:
Due to a slowdown in the industry, a Silicon Valley computer company
finds itself in debt $2,400,000. Use absolute value notation to describe
this company’s debt.
Solution:
—§ | —2,400,000 |
Exercise:
Problem:
A particular machine is set correctly if upon action its meter reads 0.
One particular machine has a meter reading of —1.6 upon action. How
far is this machine off its correct setting?
Exercises for Review
Exercise:
Problem: ((link]) Find the sum: 2 + 3+ 4.
Solution:
9
10
Exercise:
4
a
Problem: ((link]) Find the value of EUR
20
Exercise:
Problem: ({link]) Convert 3.22 to a fraction.
Solution:
13 __ 163
3 50 or 50
Exercise:
Problem:
({link]) The ratio of acid to water in a solution is How many mL of
acid are there in a solution that contain 112 mL of water?
Exercise:
Problem: ({link]) Find the value of —6 — (—8).
Solution:
2
Addition of Signed Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to add signed numbers. By
the end of the module students should be able to add numbers with like
signs and with unlike signs and be able to use the calculator for addition of
signed numbers.
Section Overview
e Addition of Numbers with Like Signs
e Addition with Zero
e Addition of Numbers with Unlike Signs
e Calculators
Addition of Numbers with Like Signs
The addition of the two positive numbers 2 and 3 is performed on the
number line as follows.
Begin at 0, the origin.
Since 2 is positive, move 2 units to the right.
Since 3 is positive, move 3 more units to the right.
We are now located at 5.
a+ SS
-2-1 01232 8 4 6 6 7
Thus, 2+3=5.
Summarizing, we have
(2 positive units) + (3 positive units) = (5 positive units)
The addition of the two negative numbers -2 and -3 is performed on the
number line as follows.
Begin at 0, the origin.
Since -2 is negative, move 2 units to the left.
Since -3 is negative, move 3 more units to the left.
We are now located at -5.
a
-7 -6 -5 -4 -3 -2 -1 0 1 2
Thus, (—2). +(—3)=—5.
Summarizing, we have
(2 negative units) + (3 negative units) = (5 negative units)
Observing these two examples, we can suggest these relationships:
(positive number) + (positive number) = (positive number)
(negative number) + (negative number) = (negative number)
Adding Numbers with the Same Sign
Addition of numbers with like sign:
To add two real numbers that have the same sign, add the absolute values of
the numbers and associate with the sum the common sign.
Sample Set A
Find the sums.
Example:
34+7
|3| 3
7 = Add these absolute values.
3+ 7=10
The common sign is “+.”
Sorte — rt OOM aa te ol:
Example:
ay)
aay =
|-9|
4+9=13
The common sign is “—.
Thus, (—4) + (—9) = —13.
Add these absolute values.
cc
Practice Set A
Find the sums.
Exercise:
Problem: 8 + 6
Solution:
14
Exercise:
Problem: 41 + 11
Solution:
DZ
Exercise:
Problem:
Solution:
-12
Exercise:
Problem:
Solution:
-45
Exercise:
Problem
: —14+ (—20)
Solution:
-34
Exercise:
Problem:
Solution:
w|~
Exercise:
Problem
: —2.8 + (—4.6)
Solution:
—T7.4
Exercise:
Problem: 0 + (—16)
Solution:
—16
Addition With Zero
Addition with Zero
Notice that
(0) + (a positive number) = (that same positive number).
(0) + (a negative number) = (that same negative number).
The Additive Identity Is Zero
Since adding zero to a real number leaves that number unchanged, zero is
called the additive identity.
Addition of Numbers with Unlike Signs
The addition 2 + (—6), two numbers with unlike signs, can also be
illustrated using the number line.
Begin at 0, the origin.
Since 2 is positive, move 2 units to the right.
Since -6 is negative, move, from 2, 6 units to the left.
We are now located at -4.
-5 -4-3-2-1 0 1 2 3 4
We can suggest a rule for adding two numbers that have unlike signs by
noting that if the signs are disregarded, 4 can be obtained by subtracting 2
from 6. But 2 and 6 are precisely the absolute values of 2 and -6. Also,
notice that the sign of the number with the larger absolute value is negative
and that the sign of the resulting sum is negative.
Adding Numbers with Unlike Signs
Addition of numbers with unlike signs: To add two real numbers that have
unlike signs, subtract the smaller absolute value from the larger absolute
value and associate with this difference the sign of the number with the
larger absolute value.
Sample Set B
Find the following sums.
Example:
7 + (—2)
7) =7 |-2| =2
Larger absolute Smaller absolute
value. Sign is positive. value.
Subtract absolute values: 7 — 2 = 5.
Attach the proper sign: "+."
Thus, 7 + (—2) = +5 or 7 + (—2) =5.
Example:
3+ (-11)
(33 Eee elnl
Smaller absolute Larger absolute
value. value. Sign is negative.
Subtract absolute values: 11 — 3 = 8.
Attach the proper sign: "—."
Thus, 3 + (—11) = —8.
Example:
The morning temperature on a winter's day in Lake Tahoe was -12 degrees.
The afternoon temperature was 25 degrees warmer. What was the
afternoon temperature?
We need to find —12 + 25.
|—12| = 12 25) = 25
Smaller absolute Larger absolute
value. value. Sign is positive.
Subtract absolute values: 25 — 12 = 16.
Attach the proper sign: "+."
Thus, —12 + 25 = 13.
Practice Set B
Find the sums.
Exercise:
Problem: 4 + (—3)
Solution:
1
Exercise:
Problem: —3 + 5
Solution:
2
Exercise:
Problem: 15 + (—18)
Solution:
-3
Exercise:
Problem: 0 + (—6)
Solution:
-6
Exercise:
Problem: —26 + 12
Solution:
-14
Exercise:
Problem: 35 + (—78)
Solution:
-43
Exercise:
Problem: 15 + (—10)
Solution:
rs)
Exercise:
Problem
:1.5+ (—2)
Solution:
-0.5
Exercise:
Problem
>:—-8+0
Solution:
-8
Exercise:
Problem
5 0+-.(0.57)
Solution:
0.57
Exercise:
Problem
: —879 + 454
Solution:
-425
Calculators
Calculators having the
+/—
key can be used for finding sums of signed numbers.
Sample Set C
Use a calculator to find the sum of -147 and 84.
Display
Reads
Type 147 147
This key changes the sign of a
Press = “147 number. It is different than —.
Press ate -147
Type 84 84
Press = -63
Practice Set C
Use a calculator to find each sum.
Exercise:
Problem: 673 + (—721)
Solution:
-48
Exercise:
Problem: —8,261 + 2,206
Solution:
-6,085
Exercise:
Problem: — 1,345.6 + (—6,648.1)
Solution:
-7,993.7
Exercises
Find the sums in the following 27 problems. If possible, use a calculator to
check each result.
Exercise:
Problem: 4 + 12
Solution:
16
Exercise:
Problem:
Exercise:
Problem:
3-0
Solution:
-15
Exercise:
Problem:
Exercise:
Problem
> 10+ (-2)
Solution:
8
Exercise:
Problem:
Exercise:
Problem
8 + (—15)
: —16+ (-9)
Solution:
-25
Exercise:
Problem
: —22 + (-1)
(—3) + (-12)
(—6) + (—20)
Exercise:
Problem:
0+ (—12)
Solution:
-12
Exercise:
Problem:
Exercise:
Problem
0+ (—4)
> 0+ (24)
Solution:
24
Exercise:
Problem
Exercise:
Problem
:—6+1+(-7)
: —5 + (—12) + (—4)
Solution:
-21
Exercise:
Problem
Exercise:
>-5+5
Problem:
Solution:
0
Exercise:
Problem:
Exercise:
Problem:
Solution:
0
Exercise:
Problem:
Exercise:
Problem:
Solution:
23
Exercise:
Problem:
Exercise:
Problem:
et en at
—14+4 14
4+(-4)
9+ (-9)
84 + (—61)
13 + (—56)
452 + (—124)
Solution:
328
Exercise:
Problem:
Exercise:
Problem:
Solution:
876
Exercise:
Problem:
Exercise:
Problem:
Solution:
-1,255
Exercise:
Problem:
Exercise:
Problem:
Solution:
-6.084
636 + (—989)
1,811 + (—935)
~373 + (—14)
—1,211 + (—44)
—47.03 + (—22.71)
—1.998 + (—4.086)
Exercise:
Problem:
In order for a small business to break even on a project, it must have
sales of $21,000. If the amount of sales was $15,000, by how much
money did this company fall short?
Exercise:
Problem:
Suppose a person has $56 in his checking account. He deposits $100
into his checking account by using the automatic teller machine. He
then writes a check for $84.50. If an error causes the deposit not to be
listed into this person’s account, what is this person’s checking
balance?
Solution:
-$28.50
Exercise:
Problem:
A person borrows $7 on Monday and then $12 on Tuesday. How much
has this person borrowed?
Exercise:
Problem:
A person borrows $11 on Monday and then pays back $8 on Tuesday.
How much does this person owe?
Solution:
$3.00
Exercises for Review
Exercise:
Problem: ({link]) Find the reciprocal of 82.
Exercise:
Problem: ((link]) Find the value of Py + ie — 7
Solution:
Ba
36
Exercise:
Problem: ({link]) Round 0.01628 to the nearest tenth.
Exercise:
Problem: ({link]) Convert 62% to a fraction.
Solution:
62 _ 31
100 ~ 50
Exercise:
Problem: (({link]) Find the value of | —12 |.
Subtraction of Signed Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to subtract signed numbers.
By the end of the module students should understand the definition of
subtraction, be able to subtract signed numbers and be able to use a
calculator to subtract signed numbers.
Section Overview
e Definition of Subtraction
e The Process of Subtraction
e Calculators
Definition of Subtraction
We know from experience with arithmetic that the subtraction 5 — 2
produces 3, that is 5 — 2 = 3. We can suggest a rule for subtracting signed
numbers by illustrating this process on the number line.
Begin at 0, the origin.
Since 5 is positive, move 5 units to the right.
Then, move 2 units to the left to get to 6. (This reminds us of addition with
a negative number.)
From this illustration we can see that 5 — 2 is the same as 5 + (—2). This
leads us directly to the definition of subtraction.
Definition of Subtraction
If a and b are real numbers, a — b is the same as a + (—b), where —0 is the
opposite of b.
The Process of Subtraction
From this definition, we suggest the following rule for subtracting signed
numbers.
Subtraction of Signed Numbers
To perform the subtraction a — 6, add the opposite of b to a, that is, change
the sign of b and add.
Sample Set A
Perform the indicated subtractions.
Example:
5—-3=5+(-3) =2
Example:
4—9=4+(-9)=-5
Example:
—4—6=-4+(-6) = -10
Example:
—3 — (-12) = -3+12=9
Example:
0 —(—15) =04+15=15
Example:
The high temperature today in Lake Tahoe was 26°F. The low temperature
tonight is expected to be -7°F. How many degrees is the temperature
expected to drop?
We need to find the difference between 26 and -7.
26 — (—7) = 26+ 7 = 33
Thus, the expected temperature drop is 33°F.
Example:
—6—(-5)-10 = -6+5+(-10)
= (-6+4+5)+(-10)
= == (—10)
= -ll
Practice Set A
Perform the indicated subtractions.
Exercise:
Problem: 9 — 6
Solution:
3
Exercise:
Problem: 6 — 9
Solution:
-3
Exercise:
Problem:
Solution:
-7
Exercise:
Problem
:1-—14
Solution:
-13
Exercise:
Problem
:—-8—12
Solution:
-20
Exercise:
Problem
: —21—-—6
Solution:
-27
Exercise:
Problem: —6 — (—4)
Solution:
ap
Exercise:
Problem: 8 — (—10)
Solution:
18
Exercise:
Problem: 1 — (—12)
Solution:
13
Exercise:
Problem: 86 — (—32)
Solution:
118
Exercise:
Problem: 0 — 16
Solution:
-16
Exercise:
Problem: 0 — (—16)
Solution:
16
Exercise:
Problem: 0 — (8)
Solution:
-8
Exercise:
Problem: 5 — (—5)
Solution:
10
Exercise:
Problem: 24 — [—(—24)|
Solution:
0
Calculators
Calculators can be used for subtraction of signed numbers. The most
efficient calculators are those with a
key.
Sample Set B
Use a calculator to find each difference.
Example:
3,187 — 8,719
Display Reads
Type 3187
Press 7
Type 8719
Press =
Thus, 3,187 — 8,719 = —5,532.
Example:
—156 — (—211)
Method A:
3187
3187
8719
-5932
Display Reads
Type 156 156
Press +/= -156
Type - -156
Press 211 211
Type += -211
Press = 318)
Thus, —156 — (—211) = 55.
Method B:
We manually change the subtraction to an addition and change the sign of
the number to be subtracted.
—156 — (—211) becomes —156 + 211
Display Reads
Type 156 156
Press +/- -156
Press a3 -156
Type 211 211
Press = 55
Practice Set B
Use a calculator to find each difference.
Exercise:
Problem: 44 — 315
Solution:
-271
Exercise:
Problem: 12.756 — 15.003
Solution:
-2.247
Exercise:
Problem: —31.89 — 44.17
Solution:
-76.06
Exercise:
Problem: —0.797 — (—0.615)
Solution:
-0.182
Exercises
For the following 18 problems, perform each subtraction. Use a calculator
to check each result.
Exercise:
Problem: 8 — 3
Solution:
5
Exercise:
Problem: 12 — 7
Exercise:
Problem: 5 — 6
Solution:
Es |
Exercise:
Problem: 14 — 30
Exercise:
Problem:
Solution:
-14
Exercise:
Problem
Exercise:
Problem:
Solution:
-2
Exercise:
Problem:
Exercise:
Problem:
Solution:
-6
Exercise:
Problem
Exercise:
Problem:
=] —12
-5 —(-8)
1 (28)
O15
j= (27)
Solution:
7
Exercise:
Problem
Exercise:
Problem:
: 0 — (—10)
67 — 38
Solution:
29
Exercise:
Problem:
Exercise:
Problem:
142 — 85
816 — 1140
Solution:
-324
Exercise:
Problem:
Exercise:
Problem:
105 — 421
—550 — (—121)
Solution:
-429
Exercise:
Problem: —15.016 — (4.001)
For the following 4 problems, perform the indicated operations.
Exercise:
Problem: —26 + 7 — 52
Solution:
-71
Exercise:
Problem: —15 — 21 — (—2)
Exercise:
Problem: — 104 — (—216) — (—52)
Solution:
164
Exercise:
Problem: —0.012 — (—0.111) — (0.035)
Exercise:
Problem:
When a particular machine is operating properly, its meter will read
34. If a broken bearing in the machine causes the meter reading to drop
by 45 units, what is the meter reading?
Solution:
“11
Exercise:
Problem:
The low temperature today in Denver was —4’F and the high was
—42°F. What is the temperature difference?
Exercises for Review
Exercise:
Problem: ({link]) Convert 16.02= to a decimal.
Solution:
16.022
Exercise:
Problem: ({link]) Find 4.01 of 6.2.
Exercise:
Problem: ({link]) Convert iz to a percent.
Solution:
oL.2570
Exercise:
Problem:
({link]) Use the distributive property to compute the product: 15 - 82.
Exercise:
Problem: ((link]) Find the sum: 16 + (—21).
Solution:
-5
Multiplication and Division of Signed Numbers
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to multiply and divide
signed numbers. By the end of the module students should be able to
multiply and divide signed numbers and be able to multiply and divide
signed numbers using a calculator.
Section Overview
¢ Multiplication of Signed Numbers
e Division of Signed Numbers
e Calculators
Multiplication of Signed Numbers
Let us consider first, the product of two positive numbers. Multiply: 3 - 5.
3-5means5+5+5=15
This suggests| footnote] that
In later mathematics courses, the word "suggests" turns into the word
"proof." One example does not prove a claim. Mathematical proofs are
constructed to validate a claim for all possible cases.
(positive number) - (positive number) = (positive number)
More briefly,
Now consider the product of a positive number and a negative number.
Multiply: (3)(—5).
(3)(—5) means (—5) + (—5) + (—5) = —15
This suggests that
(positive number) - (negative number) = (negative number)
More briefly,
Cala) =k)
By the commutative property of multiplication, we get
(negative number) - (positive number) = (negative number)
More briefly,
Jeri]
The sign of the product of two negative numbers can be suggested after
observing the following illustration.
Multiply -2 by, respectively, 4, 3, 2, 1, 0, -1, -2, -3, -4.
When this number this product increases
decreases by 1, by 2.
{
4(—2) =-8 As we know,
3(— 2) mG to ere
2(—2) | AHO) = C)
1(—2) =—2 As we know,
0(— 2) = 0—(0) - (any number) = 0
7172) = 2) the patt ted i
~9(-2) me" e pattern suggested is
— —_ — 6
CU) OH
We have the following rules for multiplying signed numbers.
Rules for Multiplying Signed Numbers
Multiplying signed numbers:
1. To multiply two real numbers that have the same sign, multiply their
absolute values. The product is positive.
(+)(+) = (+)
(=)(=) =a)
2. To multiply two real numbers that have opposite signs, multiply their
absolute values. The product is negative.
Th)
Sample Set A
Find the following products.
Example:
8-6
8] = 8
6 6 Multiply these absolute values.
8-6 = 48
Since the numbers have the same sign, the product is positive.
Thus, 8 -6=+48, or 8-6 = 48.
Example:
(—8)(—6)
[| os
6 6 Multiply these absolute values.
8-6 = 48
Since the numbers have the same sign, the product is positive.
Thus, (—8)(—6)=+48, or (—8)(—6) = 48.
Example:
(—4)(7)
Zi
\7|
4-7=28
Since the numbers have opposite signs, the product is negative.
Thus, (—4)(7) = —28.
4
i Multiply these absolute values.
Example:
6(—3)
\6|
|-3|
Os
Since the numbers have opposite signs, the product is negative.
Thus, 6(—3) = —18.
= 6
Pe) 4 Multiply these absolute values.
Practice Set A
Find the following products.
Exercise:
Problem: 3(—8)
Solution:
-24
Exercise:
Problem: 4(16)
Solution:
64
Exercise:
Problem: (—6)(—5)
Solution:
30
Exercise:
Problem: (—7)(—2)
Solution:
14
Exercise:
Problem: (—1)(4)
Solution:
-4
Exercise:
Problem: (—7)7
Solution:
-49
Division of Signed Numbers
To determine the signs in a division problem, recall that
=e = A-since 1? = 3-4
This suggests that
oa
ie = (+)
+
{= (4) since (+) = (4)(4)
What is —> + ?
—12 = (—3)(—4) suggests that 2 = —4. That is,
oe
(+) = (—)(—) suggests that Hs = (-)
What is => — Sat
—12 = (3)(—4) suggests that 32 = —4. That is,
—) = (4+)(—) suggests that iy = (-)
“——
(—)(+) suggests that > = (+4)
—
a
|
We have the following rules for dividing signed numbers.
Rules for Dividing Signed Numbers
Dividing signed numbers:
1. To divide two real numbers that have the same sign, divide their
absolute values. The quotient is positive.
Bw H =
2. To divide two real numbers that have opposite signs, divide their
absolute values. The quotient is negative.
a oO
Sample Set B
Find the following quotients.
10
9 } Divide these absolute values.
Since the numbers have opposite signs, the quotient is negative.
Thus = iy
35
7 } Divide these absolute values.
Since the numbers have the same signs, the quotient is positive.
Thus, =2 = 5.
Example:
18,
—9
Red
[18] ‘| Divide these absolute values.
Since the numbers have opposite signs, the quotient is negative.
Thus, 4 —,
Practice Set B
Find the following quotients.
Exercise:
Problem: “3h
Solution:
4
Exercise:
Problem: 2.
Solution:
-6
Exercise:
. —04
Problem: 27
Solution:
-2
Exercise:
; Bl
Problem: 17
Solution:
3
Sample Set C
Example:
; 6(4—7) —2(8—9)
Find the value of =o se
Using the order of operations and what we know about signed numbers, we
get,
RSP =270=9) S02 GT)
—(4+1)+1 _ —(5)+1
_ 1842
—5+1
= mfrac
= = 8
Practice Set C
Exercise:
5(2—6)—4(—8—1)
Problem: Find the value of 3(3-10)—9(—2)
Solution:
14
Calculators
Calculators with the
+/—
key can be used for multiplying and dividing signed numbers.
Sample Set D
Use a calculator to find each quotient or product.
Example:
(—186) - (—43)
Since this product involves a (negative) - (negative), we know the result
should be a positive number. We'll illustrate this on the calculator.
Display Reads
Type 186 186
Press f= -186
Press x -186
Type 43 43
Press +/— -43
Press = 7998
Thus, (—186) . (—43) =a dos.
Example:
— - . Round to one decimal place.
Display Reads
Type 158.64 158.64
Press Be 158.64
Type 54.3 54.3
Press Sy -54.3
Press = -2.921546961
Rounding to one decimal place we get -2.9.
Practice Set D
Use a calculator to find each value.
Exercise:
Problem: (—51.3) - (—21.6)
Solution:
1,108.08
Exercise:
Problem: —2.5746 + —2.1
Solution:
1.226
Exercise:
Problem: (0.006) - (—0.241). Round to three decimal places.
Solution:
-0.001
Exercises
Find the value of each of the following. Use a calculator to check each
result.
Exercise:
Problem: (—2)(—8)
Solution:
16
Exercise:
Problem:
Exercise:
Problem:
Solution:
32
Exercise:
Problem:
Exercise:
Problem:
Solution:
-36
Exercise:
Problem:
Exercise:
Problem:
Solution:
-60
Exercise:
(4)(-18)
(10) (—6)
Problem:
Exercise:
Problem:
Solution:
-12
Exercise:
Problem:
Exercise:
Problem:
Solution:
3
Exercise:
Problem:
Exercise:
Problem:
Solution:
-13
Exercise:
Problem:
—20
Exercise:
Problem: —
Solution:
9
Exercise:
Problem: —
Exercise:
Problem: >
Solution:
5
Exercise:
Problem: ——
Exercise:
Problem: 8 — (—3)
Solution:
11
Exercise:
Problem: 14 — (—20)
Exercise:
Problem: 20 — (—8)
Solution:
28
Exercise:
Problem: —4 — (—1)
Exercise:
Problem: 0 — 4
Solution:
-4
Exercise:
Problem: 0 — (—1)
Exercise:
Problem: —6 + 1 — 7
Solution:
-12
Exercise:
Problem: 15 — 12 — 20
Exercise:
Problem: 1 —6—7+8
Solution:
-4
Exercise:
Problem
Exercise:
Problem:
:2+7-10+2
3(4 — 6)
Solution:
-6
Exercise:
Problem:
Exercise:
Problem:
8(5 — 12)
—3(1— 6)
Solution:
ils:
Exercise:
Problem:
Exercise:
Problem:
a8(419)2
SA iB) 300 = 3)
Solution:
49
Exercise:
Problem: —9(0 — 2) + 4(8 — 9) + 0(—3)
Exercise:
Problem: 6(—2 — 9) — 6(2 + 9) + 4(—1 — 1)
Solution:
-140
Exercise:
Problem:
Exercise:
Problem:
Solution:
-7
Exercise:
Problem:
Exercise:
Problem:
Solution:
-3
Exercise:
Problem: —1(4 + 2)
Exercise:
Problem: —1(6 — 1)
Solution:
25
Exercise:
Problem: — (8 + 21)
Exercise:
Problem: —(8 — 21)
Solution:
13
Exercises for Review
Exercise:
Problem:
([link]) Use the order of operations to simplify (5? + 3? + 2)+2?.
Exercise:
Problem: ((link]) Find Sof =.
Solution:
4_ 41
3 1 3
Exercise:
Problem:
({link]) Write this number in decimal form using digits: “fifty-two
three-thousandths”
Exercise:
Problem:
({link]) The ratio of chlorine to water in a solution is 2 to 7. How many
mL of water are in a solution that contains 15 mL of chlorine?
Solution:
1
525
Exercise:
Problem: (({link]) Perform the subtraction —8 — (—20)
Summary of Key Concepts
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module reviews the key concepts from the chapter
"Signed Numbers."
Summary of Key Concepts
Variables and Constants ({link])
A variable is a letter or symbol that represents any member of a set of two
or more numbers. A constant is a letter or symbol that represents a specific
number. For example, the Greek letter 7 (pi) represents the constant
8s: eee
The Real Number Line ((link])
The real number line allows us to visually display some of the numbers in
which we are interested.
-3 -2-1 0 1 2 8
Coordinate and Graph ([link])
The number associated with a point on the number line is called the
coordinate of the point. The point associated with a number is called the
graph of the number.
Real Number ((link])
A real number is any number that is the coordinate of a point on the real
number line.
Types of Real Numbers ({link])
The set of real numbers has many subsets. The ones of most interest to us
are:
The natural numbers: {1, 2, 3, 4,...}
The whole numbers: {0, 1, 2, 3, 4,...}
The integers: {... ,-3,-2,-1,0, 1, 2, 3,...}
The rational numbers: {All numbers that can be expressed as the quotient
of two integers. }
Positive and Negative Numbers ((link])
A number is denoted as positive if it is directly preceded by a plus sign (+)
or no sign at all. A number is denoted as negative if it is directly preceded
by a minus sign (-).
Opposites ({link])
Opposites are numbers that are the same distance from zero on the number
line but have opposite signs. The numbers a and —a are opposites.
Double-Negative Property ((link])
—(-a)=a
Absolute Value (Geometric) ({link])
The absolute value of a number a, denoted | a
on the number line.
, is the distance from a to 0
Absolute Value (Algebraic) ({link])
ais a, ifa>0
a= —a, ifa<0
Addition of Signed Numbers ([link])
To add two numbers with
1. like signs, add the absolute values of the numbers and associate with
the sum the common sign.
2. unlike signs, subtract the smaller absolute value from the larger
absolute value and associate with the difference the sign of the larger
absolute value.
Addition with Zero ({link])
0 + (any number) = that particular number.
Additive Identity ({link])
Since adding 0 to any real number leaves that number unchanged, 0 is
called the additive identity.
Definition of Subtraction ({link])
a—b=a+(-b)
Subtraction of Signed Numbers ((link])
To perform the subtraction a — 6, add the opposite of b to a, that is,
change the sign of b and follow the addition rules ([link]).
Multiplication and Division of Signed Numbers ({link])
WWW BEN GO)
( (
WHI=A GHW B=EO)
( (
Exercise Supplement
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is an exercise supplement for the chapter
"Signed Numbers" and contains many exercise problems. Odd problems are
accompanied by solutions.
Exercise Supplement
Variables, Constants, and Real Numbers ([{link])
For problems 1-5, next to each real number, note all subsets of the real
numbers to which it belongs by writing N for natural numbers, W for whole
numbers, or Z for integers. Some numbers may belong to more than one
subset.
Exercise:
Problem: 61
Solution:
N,W,Z
Exercise:
Problem: — 14
Exercise:
Problem: 0
Solution:
W,Z
Exercise:
Problem: 1
Exercise:
Problem: Write all the integers that are strictly between —4 and 3
Solution:
{—3, — 2,-1,0, 1, 2}
Exercise:
Problem:
Write all the integers that are between and including —6 and —1
For each pair of numbers in problems 7-10, write the appropriate symbol
(<, >, =) in place of the 0.
Exercise:
Problem: -5 0-1
Solution:
<
Exercise:
Problem: 0 1 2
Exercise:
Problem: -7 1 0
Solution:
<
Exercise:
Problem: -1 10
For problems 11-15, what numbers can replace x so that each statement is
true?
Exercise:
Problem: —5 < x < —1, x is an integer
Solution:
{—5, —4, —3,-—2,-—1}
Exercise:
Problem: —10 < x < 0, xis a whole number.
Exercise:
Problem: 0 < x < 5, xis a natural number.
Solution:
fi. ae A}
Exercise:
Problem: —3 < xz < 3, x isa natural number
Exercise:
Problem: —8 < x < —2, x is a whole number.
Solution:
none
For problems 16-20, how many units are there between the given pair of
numbers?
Exercise:
Problem: 0 and 4
Exercise:
Problem: —1 and 3
Solution:
4
Exercise:
Problem: —7 and —4
Exercise:
Problem: —6 and 0
Solution:
6
Exercise:
Problem: —1 and 1
Exercise:
Problem:
A number is positive if it is directly preceded by a sign or no sign at
all.
Solution:
+ (plus)
Exercise:
Problem: A number is negative if it is directly preceded by a sign.
Signed Numbers ({link])
For problems 23-26, how should each number be read?
Exercise:
Problem: —8
Solution:
negative eight
Exercise:
Problem: —(—4)
Exercise:
Problem: —(—1)
Solution:
negative negative one or opposite negative one
Exercise:
Problem: —2
For problems 27-31, write each expression in words.
Exercise:
Problem: 1 + (—7)
Solution:
one plus negative seven
Exercise:
Problem: —2 — (—6)
Exercise:
Problem: —1 — (+4)
Solution:
negative one minus four
Exercise:
Problem: —(—(—3))
Exercise:
Problem: 0 — (—11)
Solution:
zero minus negative eleven
For problems 32-36, rewrite each expression in simpler form.
Exercise:
Problem: —(—4)
Exercise:
Problem: —(—15)
Solution:
15
Exercise:
Problem: —|—(—7)|
Exercise:
Problem: 1 — (—18)
Solution:
190r1+18
Exercise:
Problem: 0 — (—1)
Absolute Value ({link])
For problems 37-52, determine each value.
Exercise:
Problem: | 9 |
Solution:
a
Exercise:
Problem: | 16 |
Exercise:
Problem: | —5 |
Solution:
s
Exercise:
Problem: | —8 |
Exercise:
Problem: — | —2 |
Solution:
—2
Exercise:
Problem: — | —1 |
Exercise:
Problem: —(— | 12 |)
Solution:
iQ
Exercise:
Problem: —(— | 90 |)
Exercise:
Problem: —(— | —16 |)
Solution:
16
Exercise:
Problem
Exercise:
Problem
:—(— | 01)
2
>| —4 |
Solution:
16
Exercise:
Problem
Exercise:
Problem
>| —5 |?
3
>| —2 |
Solution:
8
Exercise:
Problem:
Exercise:
Problem
-(3-4) |
po 424
Solution:
7
Exercise:
Problem
Addition, Subtraction, Multiplication and Division of Signed Numbers
>| —7|—|—10|
({link], [link], [link])
For problems 53-71, perform each operation.
Exercise:
Problem
:-6+4
Solution:
—2
Exercise:
Problem
Exercise:
Problem:
:—-10+8
=1=6
Solution:
=f
Exercise:
Problem
Exercise:
:8—12
Problem
:0—14
Solution:
—14
Exercise:
Problem:
Exercise:
Problem:
Solution:
48
Exercise:
Problem:
Exercise:
Problem:
Solution:
—42
Exercise:
Problem:
Exercise:
Problem
5 - (—70)
: —18 + -—6
Solution:
.)
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
Exercise:
Problem:
Exercise:
Problem:
Solution:
72 + —12
—16 + —16
US==—s
—5+0
not defined
Exercise:
Problem:
Exercise:
Problem:
Solution:
—4
Exercise:
Problem:
Exercise:
Solution:
—22
—120
—|2|
Proficiency Exam
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is a proficiency exam to the chapter
"Signed Numbers." Each problem is accompanied with a reference link
pointing back to the module that discusses the type of problem
demonstrated in the question. The problems in this exam are accompanied
by solutions.
Proficiency Exam
Exercise:
Problem:
({link]) Write all integers that are strictly between —8 and —3.
Solution:
{—7, —6,-—5,- 4}
Exercise:
Problem:
({link]) Write all integers that are between and including —2 and 1.
Solution:
{—2, — 1,0, 1}
For problems 3-5, write the appropriate symbol (<, >, =) in place of the 0
for each pair of numbers.
Exercise:
Problem: ({link]) —1 0-1
Solution:
Exercise:
Problem: (({link]) 003
Solution:
<
Exercise:
Problem: ({link]) —1 0-2
Solution:
>
For problems 6 and 7, what numbers can replace z so that the statement is
true?
Exercise:
Problem: ({link]) —3 < x < 0, z is an integer.
Solution:
{—3, —2,- 1}
Exercise:
Problem: ({link]) —4 < x < 0, z is a natural number.
Solution:
{1, 2}
Exercise:
Problem: ({link]) How many units are there between —3 and 2?
Solution:
is)
For problems 9-20, find each value.
Exercise:
Problem: (({link]) | —16 |
Solution:
16
Exercise:
Problem: (({link]) — | —2 |
Solution:
—2
Exercise:
Problem: ((link]) —(—| —4 |”)
Solution:
16
Exercise:
Problem: ((link]) | —5 | + | —10 |
Solution:
15
Exercise:
Problem: ({link]) —8 + 6
Solution:
—2
Exercise:
Problem: ((link]) —3 + (—8)
Solution:
—11
Exercise:
Problem: ({link]) 0 — 16
Solution:
—16
Exercise:
Problem: ({link]) (—14) - (—3)
Solution:
42
Exercise:
Problem: ({link]) (—5 — 6)”
Solution:
121
Exercise:
Problem: ((link]) (—51) + (—7)
Solution:
2
— or? —
7 v4
Exercise:
; —4?2
Problem: ({link}) =
Solution:
6
Exercise:
—32 —15—5
Problem: ((link)) | rr |
Solution:
0)
Objectives
This module contains the learning objectives for the chapter "Algebraic
Expressions and Equations" from Fundamentals of Mathematics by Denny
Burzynski and Wade Ellis, jr.
After completing this chapter, you should
Algebraic Expressions ({link])
e be able to recognize an algebraic expression
e be able to distinguish between terms and factors
e understand the meaning and function of coefficients
e be able to perform numerical evaluation
Combining Like Terms Using Addition and Subtraction ({link]})
e be able to combine like terms in an algebraic expression
Solving Equations of the Form z + a = band z — a = b ([link])
e understand the meaning and function of an equation
e understand what is meant by the solution to an equation
e be able to solve equations of the form x + a = bandz —a=b
Solving Equations of the Form az = b and ae b ({link])
a
¢ be familiar with the multiplication/division property of equality
e be able to solve equations of the form ax = b and — = b
a
e be able to use combined techniques to solve equations
Applications I: Translating Words to Mathematical Symbols ({link])
e be able to translate phrases and statements to mathematical expressions
and equations
Applications II: Solving Problems ({link])
¢ be more familiar with the five-step method for solving applied
problems
e be able to use the five-step method to solve number problems and
geometry problems
Algebraic Expressions
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module
discusses algebraic expressions. By the end of the module students should be able to recognize an algebraic
expression, be able to distinguish between terms and factors, understand the meaning and function of
coefficients and be able to perform numerical evaluation.
Section Overview
Algebraic Expressions
e Terms and Factors
e Coefficients
e Numerical Evaluation
Algebraic Expressions
Numerical Expression
In arithmetic, a numerical expression results when numbers are connected by arithmetic operation signs (+, -,
*, +). For example, 8 + 5, 4 — 9, 3- 8, and 9 + 7 are numerical expressions.
Algebraic Expression
In algebra, letters are used to represent numbers, and an algebraic expression results when an arithmetic
operation sign associates a letter with a number or a letter with a letter. For example, z + 8,4—y,3-2,
“x+7,and «- yare algebraic expressions.
Expressions
Numerical expressions and algebraic expressions are often referred to simply as expressions.
Terms and Factors
In algebra, it is extremely important to be able to distinguish between terms and factors.
Distinction Between Terms and Factors
Terms are parts of sums and are therefore connected by + signs.
Factors are parts of products and are therefore separated by « signs.
Note: While making the distinction between sums and products, we must remember that subtraction and
division are functions of these operations.
1. In some expressions it will appear that terms are separated by minus signs. We must keep in mind that
subtraction is addition of the opposite, that is,
t—y=2x+(—y)
2. In some expressions it will appear that factors are separated by division signs. We must keep in mind that
1
z xz 1
or . — xz .
ae y
Sample Set A
State the number of terms in each expression and name them.
Example:
x + 4. In this expression, x and 4 are connected by a'"+" sign. Therefore, they are terms. This expression
consists of two terms.
Example:
y — 8. The expression y — 8 can be expressed as y + (—8). We can now see that this expression consists of
the two terms y and —8.
Rather than rewriting the expression when a subtraction occurs, we can identify terms more quickly by
associating the + or - sign with the individual quantity.
Example:
a+7-—b-—~m. Associating the sign with the individual quantities, we see that this expression consists of the
four terms a, 7, —b, —m.
Example:
5m — 8n. This expression consists of the two terms, 5m and —8n. Notice that the term 5m is composed of
the two factors 5 and m. The term —8n is composed of the two factors —8 and n.
Example:
32. This expression consists of one term. Notice that 32 can be expressed as 3a + 0 or 3z - 1 (indicating the
connecting signs of arithmetic). Note that no operation sign is necessary for multiplication.
Practice Set A
Specify the terms in each expression.
Exercise:
Problem: x + 7
Solution:
x, 7
Exercise:
Problem: 3m — 6n
Solution:
3m — 6n
Exercise:
Problem: 5y
Solution:
Sy
Exercise:
Problem: a + 2b —c
Solution:
a, 2b, —c
Exercise:
Problem: —3xz — 5
Solution:
—32,—5
Coefficients
We know that multiplication is a description of repeated addition. For example,
5:7 describess7+7+7+7+7
Suppose some quantity is represented by the letter z. The multiplication 5x describes 7 + x2+2+2+442. Itis
now easy to see that 5x specifies 5 of the quantities represented by a. In the expression 52, 5 is called the
numerical coefficient, or more simply, the coefficient of x.
Coefficient
The coefficient of a quantity records how many of that quantity there are.
Since constants alone do not record the number of some quantity, they are not usually considered as numerical
coefficients. For example, in the expression 7x + 2y — 8z + 12, the coefficient of
7x is 7. (There are 7 x's.)
2y is 2. (There are 2 y's.)
—8zis —8. (There are —8z's.)
The constant 12 is not considered a numerical coefficient.
lz=<2z
When the numerical coefficient of a variable is 1, we write only the variable and not the coefficient. For
example, we write x rather than 12. It is clear just by looking at x that there is only one.
Numerical Evaluation
We know that a variable represents an unknown quantity. Therefore, any expression that contains a variable
represents an unknown quantity. For example, if the value of x is unknown, then the value of 3x + 5 is
unknown. The value of 3x + 5 depends on the value of z.
Numerical Evaluation
Numerical evaluation is the process of determining the numerical value of an algebraic expression by
replacing the variables in the expression with specified numbers.
Sample Set B
Find the value of each expression.
Example:
de fay it ¢ — —4 and 4 — 2
Replace x with -4 and y with 2.
2e+7y = 2(—4)+7(2)
= =p 14
6
Thus, when « =—4 and y = 2, 2a + Ty = 6.
Example:
5a 8b .
ae + Jo’ fa = 6 andb= —3.
Replace a with 6 and b with —3.
ba , 8 _ 5(6) | 8(-3)
7 eS = oo
= mfrac + mfrac
= —10+4 (-2)
= -12
b
Thus, when a = 6 and b =-3, oa + 5 = —12.
b iy
Example:
6(2a — 15b), ifa = —5 andb = -1
Replace a with —5 and 6 with -1.
6(2a — 156) = 6(2(—5) — 15(-1))
= 6(—10 + 15)
= 6(5)
30
Thus, when a = —5 and b = -1, 6(2a — 15b) = 30.
Example:
82” —2e4+1,ife=—4
Replace x with 4.
322 -—2¢+1 = 3(4)? — 2(4) aT
= 3-16-—2(4)+1
= 43 = il
= dil
Thus, when x = 4, 3a? — 22 +1=41.
Example:
-27? —Aifx=3
Replace x with 3.
—¢?—-4 = —3 —4 Be careful to square only the 3. The exponent 2 is connected only to 3, not -3
= =)=4
= —13
Example:
(—2) —4,ifx = 3.
Replace x with 3.
(-2)’-4 = (-3
= $4
—5
The exponent is connected to —3, not 3 as in the problem above.
a Awl he exponent is connected to -3, not 3 as in problem 5 above.
Practice Set B
Find the value of each expression.
Exercise:
Problem: 9m — 2n, ifm = —2 andn = 5
Solution:
-28
Exercise:
Problem: —3z — 5y+ 2z, ifx = —4,y=3,z=0
Solution:
3
Exercise:
1 4b
Problem: ~ ae ifa = —6, andb= 2
Solution:
-6
Exercise:
Problem: 8(3m — 5n), ifm = —4andn = —5
Solution:
104
Exercise:
Problem: 3/—40 — 2(4a — 3b)], ifa = —6 andb =0
Solution:
24
Exercise:
Problem: by? + 6y —11,ify=—-1
Solution:
-12
Exercise:
Problem: —x? + 2x + Tike 4
Solution:
-1
Exercise:
Problem: (—«)” + 2x +7, if2 = 4
Solution:
31
Exercises
Exercise:
Problem: In an algebraic expression, terms are separated by signs and factors are separated by signs.
Solution:
Addition; multiplication
For the following 8 problems, specify each term.
Exercise:
Problem: 3m + 7n
Exercise:
Problem: 5x + 18y
Solution:
5x ,18y
Exercise:
Problem: 4a — 6b +c
Exercise:
Problem: 8s + 2r — 7t
Solution:
8s,2r, — Tt
Exercise:
Problem: m — 3n — 4a + 7b
Exercise:
Problem: 7a — 2b — 3c — 4d
Solution:
7a, — 2b, — 3c, — 4d
Exercise:
Problem: —6a — 5b
Exercise:
Problem: —z — y
Solution:
—&@,—Y
Exercise:
Problem: What is the function of a numerical coefficient?
Exercise:
Problem: Write 1m in a simpler way.
Solution:
m
Exercise:
Problem: Write 1s in a simpler way.
Exercise:
Problem: In the expression 5a, how many a’s are indicated?
Solution:
5
Exercise:
Problem: In the expression —7c, how many c’s are indicated?
Find the value of each expression.
Exercise:
Problem: 2m — 6n, ifm = —3 andn = 4
Solution:
-30
Exercise:
Problem: 5a + 60, ifa = —6 andb=5
Exercise:
Problem: 2x2 — 3y+ 4z, if~ =1, y= —1, andz= —2
Solution:
-3
Exercise:
Problem: 9a + 6b — 8a + 4y, if a 2,6 1a
Exercise:
8 18
Problem: he + coal if =9andy= —2
3y 22x
Solution:
-14
Exercise:
—3m —6n .
Problem: ,ifm = —6andn = 3
2n m
Exercise:
Problem: 4(3r + 2s), ifr = 4 ands = 1
Solution:
56
Exercise:
2,and y = 0
Problem:
Exercise:
Problem:
Solution:
64
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
Exercise:
Problem:
Exercise:
Problem
(a — 3b + 2c — d), ifa
3(9a — 6b), if a = —1andb = —2
—8(5m + 8n), ifm = 0andn = —1
—2(-62 + y — 2z), iff =1,y=1,andz=2
—(102 — 2y+5z)ife =2,y=8, andz=-1
—5,b=2,c=0,andd=—1
: 3[16 — 3(a + 3b)|, if a = 3 and b = —2
Solution:
75
Exercise:
Problem
Exercise:
Problem:
: —2[5a + 2b(b — 6)], ifa = —2 andb=3
Solution:
24
Exercise:
Problem:
Exercise:
Problem
b
2{19 — 6/4 — 2(a
2° +32 —-1,ife=5
Solution:
{6x + 3y[—2(a + 4y)|}, ifs =Oandy=1
7)|}, ifa = 10 andb=3
39
Exercise:
Problem: m” — 2m + 6, if m = 3
Exercise:
Problem: 6a” + 2a — 15, ifa = —2
Solution:
5
Exercise:
Problem: 5s” + 65 + 10, ifz = —-1
Exercise:
Problem: 16x? + 8x2 — 7,ife=0
Solution:
-7
Exercise:
Problem: —8y” + 6y + 11, if y= 0
Exercise:
Problem: (y — 6)” + 3(y—5) +4, ify =5
Solution:
5
Exercise:
Problem: (x + 8)” + 4(z +9) +1, if2 = —6
Exercises for Review
Exercise:
: 2 3 1
Problem: ((link]) Perform the addition: aa + re
Solution:
181 13
(24 «24
Exercise:
15
11
Problem: ((link]) Arrange the numbers in order from smallest to largest: —
Exercise:
2
2
Problem: ((link]) Find the value of (5) + =
Solution:
20
27
Exercise:
Problem: ((link]) Write the proportion in fractional form: “9 is to 8 as x is to 7.”
Exercise:
Problem: (([link]) Find the value of —3(2 — 6) — 12
Solution:
0
»——,) an
32° 48
7
16
Combining Like Terms Using Addition and Subtraction
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to combine like terms using
addition and subtraction. By the end of the module students should be able
to combine like terms in an algebraic expression.
Section Overview
e¢ Combining Like Terms
Combining Like Terms
From our examination of terms in [link], we know that like terms are terms
in which the variable parts are identical. Like terms is an appropriate name
since terms with identical variable parts and different numerical coefficients
represent different amounts of the same quantity. When we are dealing with
quantities of the same type, we may combine them using addition and
subtraction.
Simplifying an Algebraic Expression
An algebraic expression may be simplified by combining like terms.
This concept is illustrated in the following examples.
1. 8 records + 5 records = 13 records.
Eight and 5 of the same type give 13 of that type. We have combined
quantities of the same type.
2.8 records + 5 records + 3 tapes = 13 records + 3 tapes.
Eight and 5 of the same type give 13 of that type. Thus, we have 13 of
one type and 3 of another type. We have combined only quantities of
the same type.
3. Suppose we let the letter z represent "record." Then, 82 + 5x = 132.
The terms 8x and 5z are like terms. So, 8 and 5 of the same type give
13 of that type. We have combined like terms.
4. Suppose we let the letter x represent "record" and y represent "tape."
Then,
82 + be ++ 3y = 132+ by
We have combined only the like terms.
After observing the problems in these examples, we can suggest a method
for simplifying an algebraic expression by combining like terms.
Combining Like Terms
Like terms may be combined by adding or subtracting their coefficients and
affixing the result to the common variable.
Sample Set A
Simplify each expression by combining like terms.
Example:
2m + 6m — 4m. All three terms are alike. Combine their coefficients and
affix this result tom: 2+6—4= 4.
Thus, 2m + 6m — 4m = 4m.
Example:
ox + 2y — Oy. The terms 2y and —9y are like terms. Combine their
coefficients: 2 — 9 = —7.
Thus, 52 --2y —9y — 52 — fy.
Example:
—3a + 2b —5a+a-+ 6b. The like terms are
—3a, — 5a,a 2b, 6b
=o al —— O_o
—Ta 8b
Thus, —3a + 2b — 5a+ a+ 6b= — 7a+ 8b.
Example:
r—28s+ 7s + 3r—4r —5s. The like terms are
r, or, 49 — 2s, Ts, —58
| mad Edt Et
1+3-4=0 -2+7-5=0
Or Os
Or+ 0s =0
Thus, r— 2s + 7s+ 3r—4r—5s=0.
Practice Set A
Simplify each expression by combining like terms.
Exercise:
Problem: 4z + 3x2 + 6z
Solution:
132
Exercise:
Problem: 5a + 8)b+ 6a — 2b
Solution:
lla + 6b
Exercise:
Problem: 10m — 6n —2n —-m+n
Solution:
9m — 7n
Exercise:
Problem: 16a + 6m + 2r — 3r —18a+ m— 7m
Solution:
=260—7
Exercise:
Problem: 5h — 8k + 2h — 7h+ 3k + 5k
Solution:
0
Exercises
Simplify each expression by combining like terms.
Exercise:
Problem: 4a + 7a
Solution:
lla
Exercise:
Problem: 3m + 5m
Exercise:
Problem:
Solution:
Ah
Exercise:
Problem:
Exercise:
Problem:
Solution:
3m + 3n
Exercise:
Problem:
Exercise:
Problem:
Solution:
17s—r
Exercise:
Problem:
Exercise:
Problem:
6h — 2h
11k — 8k
5m + 3n — 2m
(2 — O62 + 3y
14s + 3s — 8r+7r
—5m — 3n+ 2m-+ 6n
7Th+ 3a —10k+ 6a — 2h — 5k — 3k
Solution:
5h + 9a — 18k
Exercise:
Problem: 4x7 — 8y — 32-+- «& —y—z— dy — 2z
Exercise:
Problem: llw + 32 — 6w — 5w-+ 8a — 1lz
Solution:
0
Exercise:
Problem: 15r — 6s + 2r+ 8s — 6r — 7s —s —2r
Exercise:
Problem: | —7 | m+ | 6 | m+ | —3|m
Solution:
16m
Exercise:
Problem: | —2 | z+ | —8| z+ | 10| az
Exercise:
Problem: (—4 + 1)k+ (6 — 3)k+ (12 —4)h+ (54+ 2)k
Solution:
8h + 7k
Exercise:
Problem: (—5 + 3)a — (2+5)b— (3+ 8)b
Exercise:
Problem: 5« + 2A + 3A — 8x
Solution:
5A — 3x
Exercise:
Problem: 9X + 10H — 11 — 12H
Exercise:
Problem: 16z — 12y+ 5x + 7 — 5x2 — 16 — 3y
Solution:
l6z — 1by — 9
Exercise:
Problem: —3y + 4z — 11 — 3z — 2y+ 5 — 4(8 — 3)
Exercises for Review
Exercise:
24
Problem: ({link]) Convert aa to a mixed number
Solution:
2
11
Exercise:
3
Problem: ([link]) Determine the missing numerator: 8 64°
Exercise:
SL
6 4
Problem: ({link]) Simplify i
12
Solution:
7
Exercise:
5
Problem: ({link]) Convert 16 to a percent.
Exercise:
Problem: ({link]) In the expression 6k, how many k’s are there?
Solution:
6
Solving Equations of the Form x+a=b and x-a=b
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis,
Jr. This module discusses how to solve equations of the form x + a = band « —a = b.
By the end of the module students should understand the meaning and function of an
equation, understand what is meant by the solution to an equation and be able to solve
equations of the form z+ a= bandz—a=b.
Section Overview
e Equations
¢ Solutions and Equivalent Equations
¢ Solving Equations
Equations
Equation
An equation is a statement that two algebraic expressions are equal.
The following are examples of equations:
zx+6 = 10 x-4 = -11 3y-—5 = -2+42y
This This This This This This
expression = expression expression = expression expression = expression
Notice that z + 6, x — 4, and 3y — 5 are not equations. They are expressions. They are
not equations because there is no statement that each of these expressions is equal to
another expression.
Solutions and Equivalent Equations
Conditional Equations
The truth of some equations is conditional upon the value chosen for the variable. Such
equations are called conditional equations. There are two additional types of equations.
They are examined in courses in algebra, so we will not consider them now.
Solutions and Solving an Equation
The set of values that, when substituted for the variables, make the equation true, are
called the solutions of the equation.
An equation has been solved when all its solutions have been found.
Sample Set A
Example:
Verify that 3 is a solution to x + 7 = 10.
When x = 3,
fee ek eS AD
becomes 3+7 = 10
10 = 10 whichisatrue statement, verifying that
3 isasolutionto «+ 7= 10
Example:
Verify that —6 is a solution to 5y + 8 = —22
When y = —6,
5y +8 = —22
becomes 5(—6)+8 = —22
—30+8 = —22
—22 = -—22 whichisatrue statement, verifying that
—6 is asolution to5y + 8 = —22
Example:
Verify that 5 is not a solution toa — 1 = 2a + 3.
When a = 5,
a-1l = 2a+3
becomes 5—1 = 2-543
5-1 = 10+3
Ane ALS a false statement, verifying that 5
is not asolution to a —1 = 2a+3
Example:
Verify that -2 is a solution to 3m — 2 = —4m — 16.
When m = —2,
3m—2 = —4m-—16
becomes 3(—2)—2 = —4(—2) —16
—6-—2 = 8-16
—-8§ = -8 which is a true statement, verifying that
—2 isasolution to 3m — 2 = —4m — 16
Practice Set A
Exercise:
Problem: Verify that 5 is a solution tom + 6 = 11.
Solution:
Substitute 5 intom + 6 = 11.
5+6211
11411
Thus, 5 is a solution.
Exercise:
Problem: Verify that —5 is a solution to 2m — 4 = —14.
Solution:
Substitute -5 into 2m — 4 = —14.
2(—5) —42—-14
—10-—42-14
—144-14
Thus, -5 is a solution.
Exercise:
Problem: Verify that 0 is a solution to 5a + 1 = 1.
Solution:
Substitute 0 into 5a +1= 1.
5(0) +121
0+121
141
Thus, 0 is a solution.
Exercise:
Problem: Verify that 3 is not a solution to —3y + 1 = 4y+ 5.
Solution:
Substitute 3 into —-3y +1 = 4y+5.
—3(3) +12 4(3) +5
—-9+1212+5
—8#17
Thus, 3 is not a solution.
Exercise:
Problem: Verify that -1 is a solution to 6m — 5+ 2m = 7m — 6.
Solution:
Substitute -1 into 6m — 5+ 2m = 7m — 6.
6(—1) —5 + 2(—-1) 2 7(-1) -6
—6—-5-22-7-6
—13 4-13
Thus, -1 is a solution.
Equivalent Equations
Some equations have precisely the same collection of solutions. Such equations are called
equivalent equations. For example, x — 5 = —1,a+7= 11, andz = 4areall
equivalent equations since the only solution to each is z = 4. (Can you verify this?)
Solving Equations
We know that the equal sign of an equation indicates that the number represented by the
expression on the left side is the same as the number represented by the expression on the
right side.
This number is the same as this number
z+i7 = 11
xr—5 = -1
Addition/Subtraction Property of Equality
From this, we can suggest the addition/subtraction property of equality.
Given any equation,
1. We can obtain an equivalent equation by adding the same number to both sides of the
equation.
2. We can obtain an equivalent equation by subtracting the same number from both
sides of the equation.
The Idea Behind Equation Solving
The idea behind equation solving is to isolate the variable on one side of the equation.
Signs of operation (+, -, ‘,+) are used to associate two numbers. For example, in the
expression 5 + 3, the numbers 5 and 3 are associated by addition. An association can be
undone by performing the opposite operation. The addition/subtraction property of
equality can be used to undo an association that is made by addition or subtraction.
Subtraction is used to undo an addition.
Addition is used to undo a subtraction.
The procedure is illustrated in the problems of [link].
Sample Set B
Use the addition/subtraction property of equality to solve each equation.
Example:
z2+4=6.
4 is associated with x by addition. Undo the association by subtracting 4 from both sides.
xz+4-4=6-4
c£+0=2
t= 2
Check: When x = 2, x + 4 becomes
2+426
626.
The solution tox + 4= 6isz = 2.
Example:
m — 8 = 5. 8 is associated with m by subtraction. Undo the association by adding 8 to
both sides.
Oo oie
m+0=13
ie = 1183
Check: When m = 13,
becomes
m—-8=5
138-825
645
a true statement.
The solution tom — 8 = 5ism = 13.
Example:
—3—5=y-—2-+ 8. Before we use the addition/subtraction property, we should
simplify as much as possible.
—3—-5=y-—2+8
a
6 is associated with y by addition. Undo the association by subtracting 6 from both sides.
Or Oe On
Si eo
—14=y
This is equivalent to y= —14.
Check: When y = —14,
—3-—5=y-2+8
becomes
—3—5L—-14—-2+8
—82~16+8
—8 2-8,
a true statement.
The solution to —3 —5 = y—2+8isy = —14.
Example:
—5a+1-+ 6a = —2. Begin by simplifying the left side of the equation.
Spe ai
—5+6=1
a+ 1 = —2 1 is associated with a by addition. Undo the association by subtracting 1
from both sides.
a+1l-1=-2-1
at+0=—3
C=
Check: When a = —3,
Soda bo Od = —=2
becomes
—5(—3) + 1+ 6(—3) 2 —2
16+1-18i-2
—-24-2
9
a true statement.
The solution to —5a + 1+ 6a = —2isa = —3.
Example:
7k — 4 = 6k + 1. In this equation, the variable appears on both sides. We need to isolate
it on one side. Although we can choose either side, it will be more convenient to choose
the side with the larger coefficient. Since 8 is greater than 6, we’ll isolate & on the left
side.
7k — 4 = 6k + 1 Since 6k represents +6k, subtract 6k from each side.
7k-—4-—-6k=6k+1-—6k
7-6=1 6—6=0
k — 4 = 1 4is associated with k by subtraction. Undo the association by adding 4 to both
sides.
k—-4+4=1+44
= 5
Check: When k = 5,
7kh-4=6k+1
becomes
7°5-416-5+1
35-42304+1
31 ~ 31.
a true statement.
The solution to 7k -4 = 6k+1isk=5.
Example:
—8 + x = 5. -8 is associated with x by addition. Undo the by subtracting -8 from both
sides. Subtracting -8 we get —(—8)=+8. We actually add 8 to both sides.
SS EHS eh = ses.
hie el
Check: When « = 13
Sila oy aad
becomes
—8+1325
5 %5
’
a true statement.
The solution to -8 +2 = 5isx = 13.
Practice Set B
Exercise:
Problem: y + 9 = 4
Solution:
a
Exercise:
Problem: a — 4 = 11
Solution:
a=15
Exercise:
Problem: —1+7=2+3
Solution:
p= 3
Exercise:
Problem: 8m + 4 — 7m = (—2)(-—3)
Solution:
m=2
Exercise:
Problem: 12k — 4 = 9k —6+4 2k
Solution:
k= —-2
Exercise:
Problem: —3 + a = —4
Solution:
a=-l
Exercises
For the following 10 problems, verify that each given value is a solution to the given
equation.
Exercise:
Problem: z — 11 = 5, x = 16
Solution:
Substitute z = 4 into the equation 4z — 11 = 5.
16-—11=5
5=5
x = 4is a solution.
Exercise:
Problem: y — 4 = —6, y = —2
Exercise:
Problem: 2m —1=1,m=1
Solution:
Substitute m = 1 into the equation 2m — 1 = 1.
2-111
1¥#1
m = 1 isa solution.
Exercise:
Problem: 5y + 6 = —14,y = —4
Exercise:
Problem: 3x + 2 — 7x = —5x — 6,2 = —8
Solution:
Substitute z = —8 into the equation 3x + 2 — 7 =
—24+2—7240-6
34 ~ 34
az = —8 is a solution.
Exercise:
Problem: —6a + 3+ 3a = 4a+ 7 — 3a,a = —1
Exercise:
Problem: —8 + z = —8,xz = 0
Solution:
Substitute z = 0 into the equation —8 + x = —8.
—5x — 6.
—-8+0i-8
—-8i-8
zx = Oisa solution.
Exercise:
Problem: 8b + 6 = 6 — 5b,b=0
Exercise:
L5
Problem: 4x — 5 = 6x — 20,7 = oa
Solution:
15
Substitute x = ae the equation 4z — 5 = 6a — 20.
30 — 5 2 45 — 20
25 ~ 25
15. :
a cy is a solution.
Exercise:
22
Problem: —3y + 7 = 2y — 15, y = =
Solve each equation. Be sure to check each result.
Exercise:
Problem: y — 6 = 5
Solution:
g=11
Exercise:
Problem: m+ 8 = 4
Exercise:
Problem: k — 1 = 4
Solution:
k=5
Exercise:
Problem: h — 9 = 1
Exercise:
Problem: a + 5 = —4
Solution:
a=-—9
Exercise:
Problem: b — 7 = —1
Exercise:
Problem: xz + 4—9=6
Solution:
z-—11
Exercise:
Problem: y — 8 + 10 = 2
Exercise:
Problem: z + 6 = 6
Solution:
z-—0
Exercise:
Problem: w — 4 = —4
Exercise:
Problem: x + 7 —9=6
Solution:
zr—8
Exercise:
Problem: y— 2+5=4
Exercise:
Problem: m + 3 — 8 = —6+2
Solution:
m=1
Exercise:
Problem: z + 10 — 8 = —8+ 10
Exercise:
Problem: 2+ 9=k-—8
Solution:
v= 19
Exercise:
Problem: —5+3=—h—4
Exercise:
Problem: 3m — 4 = 2+ 6
Solution:
m= 10
Exercise:
Problem: 5a + 6 = 4a — 8
Exercise:
Problem: 8b + 6 + 2b = 3b -—7+6b-—8
Solution:
b= -—21
Exercise:
Problem:
Exercise:
Problem:
Solution:
a=-16
Exercise:
Problem:
12h —1—3-—5h =2h+5h+3(—4)
—4a +5 —-— 2a = —3a — 11 —-2a
—9n — 2—6-+ 5n = 3n — (2)(—5) — 6n
Calculator Exercises
Exercise:
Problem: y — 2.161 = 5.063
Solution:
y = 7.224
Exercise:
Problem:
Exercise:
Problem:
Solution:
a — 44.0014 = —21.1625
—0.362 — 0.416 = 5.63m — 4.63m
m = —0.778
Exercise:
Problem: 8.078 — 9.112 = 2.106y — 1.106y
Exercise:
Problem: 4.23k + 3.18 = 3.23k — 5.83
Solution:
k = —9.01
Exercise:
Problem: 6.1185z — 4.0031 = 5.11852 — 0.0058
Exercise:
Problem: 21.63y + 12.40 — 5.09y = 6.1ly — 15.66 + 9.43y
Solution:
y = —28.06
Exercise:
Problem: 0.029a — 0.013 — 0.034 — 0.057 = —0.038 + 0.56 + 1.01a
Exercises for Review
Exercise:
T7calculators
Problem: ({link]) Is a ibeeataebe an example of a ratio or a rate?
12students
Solution:
rate
Exercise:
3
Problem: ([link]) Convert a to a decimal.
Exercise:
Problem: (({link]) 0.4% of what number is 0.014?
Solution:
3.5
Exercise:
Problem:
([link]) Use the clustering method to estimate the sum: 89 + 93 + 206 + 198 + 91
Exercise:
Problem: ({link]) Combine like terms: 4x” + 8y + 12y + 9a — 2y.
Solution:
13z + 18y
Solving Equations of the Form ax=b and x/a=b
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses solving equations of the form
ax = band ees b. By the end of the module students should be familiar
a
with the multiplication/division property of equality, be able to solve
equations of the form ax = b and — = band be able to use combined
a
techniques to solve equations.
Section Overview
¢ Multiplication/ Division Property of Equality
e Combining Techniques in Equations Solving
Multiplication/ Division Property of Equality
Recall that the equal sign of an equation indicates that the number
represented by the expression on the left side is the same as the number
represented by the expression on the right side. From this, we can suggest
the multiplication/division property of equality.
Multiplication/Division Property of Equality
Given any equation,
1. We can obtain an equivalent equation by multiplying both sides of the
equation by the same nonzero number, that is, if c ~ 0, then a = bis
equivalent to
G2C=b)ee
2. We can obtain an equivalent equation by dividing both sides of the
equation by the same nonzero number, that is, if c ~ 0, then a = bis
equivalent to
a b
The multiplication/division property of equality can be used to undo an
association with a number that multiplies or divides the variable.
Sample Set A
Use the multiplication / division property of equality to solve each equation.
Example:
6y = 54
6 is associated with y by multiplication. Undo the association by dividing
both sides by 6
Se)
6 6
By _
aa
9
Check: When y = 9
6y = 54
becomes
6-9254
54 Z 54,
5)
a true statement.
The solution to 6y = 54 is y = 9.
Example:
eee
—2
-2 is associated with x by division. Undo the association by multiplying
both sides by -2.
(22) = (-2)27
= — 54
Check: When x = —54,
ete 7
—2
becomes
— 54
= Ea tf
27 £27
a true statement.
2 x ;
The solution to = = AS —— 54
Example:
3
ae.
7
We will examine two methods for solving equations such as this one.
Method 1: Use of dividing out common factors.
aoe
ae
7 is associated with a by division. Undo the association by multiplying
both sides by 7.
3a
i SED
iG
Divide out the 7’s.
3
bee o — 42
3a = 42
3 is associated with a by multiplication. Undo the association by dviding
both sides by 3.
Check: When a = 14,
3a
enaieas
7
becomes
a true statement.
3a
The solution to 7 6:15 1 — 14,
Method 2: Use of reciprocals
Recall that if the product of two numbers is 1, the numbers are reciprocals.
Thus : and x are reciprocals.
3a
na
Multiply both sides of the equation by = the reciprocal of =
7 3a 7
SU)
Notice that we get the same solution using either method.
Example:
=§e = 24
-8 is associated with x by multiplication. Undo the association by dividing
both sides by -8.
ie pds
=e 28
—8xr 24
—8 —8
x= -3
Check: When x = —3,
—8xr = 24
becomes
—8(—3) 2 24
24 ¥ 24,
5)
a true statement.
Example:
ie i
Since — x is actually —1- 2 and (—1)(—1) = 1, we can isolate x by
multiplying both sides of the equation by —1.
(—1)(—x2) = -1-7
oe
Check: When z = 7,
oie
becomes
—(-1) 27
7L7
The solution to —x = 7is x = —7.
Practice Set A
Use the multiplication/division property of equality to solve each equation.
Be sure to check each solution.
Exercise:
Problem:
Solution:
pees
Exercise:
Problem:
Solution:
p13
Exercise:
Problem:
Solution:
xz — —32
Exercise:
Problem:
Solution:
xz—16
Exercise:
[Ce Al
—dxz = 65
Problem: —y = 3
Solution:
y=—3
Exercise:
Problem: —k = —2
Solution:
k=2
Combining Techniques in Equation Solving
Having examined solving equations using the addition/subtraction and the
multiplication/division principles of equality, we can combine these
techniques to solve more complicated equations.
When beginning to solve an equation such as 6x2 — 4 = —16, it is helpful
to know which property of equality to use first, addition/subtraction or
multiplication/division. Recalling that in equation solving we are trying to
isolate the variable (disassociate numbers from it), it is helpful to note the
following.
To associate numbers and letters, we use the order of operations.
1. Multiply/divide
2. Add/subtract
To undo an association between numbers and letters, we use the order of
operations in reverse.
1. Add/subtract
2. Multiply/divide
Sample Set B
Solve each equation. (In these example problems, we will not show the
checks.)
Example:
67 —4 => —16
-4 is associated with x by subtraction. Undo the association by adding 4 to
both sides.
Of 4-4 — 16 4
Otis 12
6 is associated with x by multiplication. Undo the association by dividing
both sides by 6
6x Sailr
6 6s «6
—
Example:
OK; o——4p:
3 is associated with k by addition. Undo the association by subtracting 3
from both sides.
—8k +3 —3=—45 —3
—8k= — 48
-8 is associated with k by multiplication. Undo the association by dividing
both sides by -8.
—8k —48
Sa. os
a6
Example:
5m — 6 — 4m = 4m — 8+ 3m. Begin by solving this equation by
combining like terms.
m — 6 = 7m — 8 Choose a side on which to isolate m. Since 7 is greater
than 1, we'll isolate m on the right side.
Subtract m from both sides.
S10 — i — fn 8 — IN
=0 — O13
8 is associated with m by subtraction. Undo the association by adding 8 to
both sides.
=O. O— Ot; et >
2=>6m
6 is associated with m by multiplication. Undo the association by dividing
both sides by 6.
aye ils Reduce
A =p uce.
1
—=-m
3
Notice that if we had chosen to isolate m on the left side of the equation
rather than the right side, we would have proceeded as follows:
m—6=7m-—8
Subtract 7m from both sides.
m—6—7m=7m-—8—7m
OU OS
Add 6 to both sides,
ON = CoO — = OO
Atte
Divide both sides by -6.
—6m 2
6 -6
i]
ase
3
This is the same result as with the previous approach.
Example:
8
8g
7
7 is associated with x by division. Undo the association by multiplying
both sides by 7.
8x
ats SG.
foe
8
eer
8 is associated with x by multiplication. Undo the association by dividing
both sides by 8.
pes os
Practice Set B
Solve each equation. Be sure to check each solution.
Exercise:
Problem: 5m + 7= — 13
Solution:
m=—A4
Exercise:
Problem: —3a — 6 = 9
Solution:
=)
Exercise:
Problem: 2a + 10 — 3a = 9
Solution:
a |
Exercise:
Problem: llz — 4— 132 = 47+ 14
Solution:
C= 3
Exercise:
Problem: —3m + 8 = —5m+1
Solution:
eS —
2
Exercise:
Problem: 5y + 8y — 11= — 11
Solution:
y=0
Exercises
Solve each equation. Be sure to check each result.
Exercise:
Problem: 7x = 42
Solution:
2r=6
Exercise:
Problem:
Exercise:
Problem:
Solution:
Pa 1D
Exercise:
Problem:
Exercise:
Problem:
Solution:
a= 3s
Exercise:
Problem:
Exercise:
Problem:
Solution:
y= 14
Exercise:
8x — 81
10z = 120
lig = 121
—6a = 48
—9y = 54
—sy=— 42
Problem: —5a=— — 105
Exercise:
Problem: 2™= — 62
Solution:
m= — 3l
Exercise:
Problem: 3m= — 54
Exercise:
Problem:
m8
Solution:
P= 25
Exercise:
Problem: — pe I |
wc
Exercise:
Problem: — = — 14
Solution:
z— 84
Exercise:
Problem:
Exercise:
Problem:
Solution:
m—-—A4
Exercise:
Problem
Exercise:
Problem
Solution:
zr—-l
Exercise:
Problem
Exercise:
Problem
ae
5
3m —1=-— 13
2:42 4+ 7=-17
:2+9r=—7
(5: = lle = 27
:32 = 4y4+ 6
Solution:
a 13
a)
Exercise:
Y
Problem: —5 + 4 = —8m + 1
Exercise:
Problem: 3k + 6 = 5k + 10
Solution:
k= —2
Exercise:
Problem: 4a + 16 = 6a + 8a+ 6
Exercise:
Problem: 6z + 5+ 2x2 —1= 92 — 32 +15
Solution:
11 1
— or 5 —
zr 5 or 5
Exercise:
Problem: —9y — 8+ 3y+ 7 = —7y+ 8y— 5y+ 9
Exercise:
Problem: —3a =—a+5
Solution:
5
a= -—
A
Exercise:
Problem:
Exercise:
Problem:
Solution:
m=
Exercise:
Problem:
Exercise:
Problem:
Solution:
r= 0
Exercise:
Problem:
Exercise:
Problem:
Solution:
GH—14
Exercise:
p0= = 2b: 8b Fl
—3m +2-—-8m-—-4=-14m+m™-4
5at+3=3
7x + 32 = 0
7g+4-—1lg=-4g+1+g
= 10
a
Problem:
Exercise:
Problem:
Solution:
eed 6)
Exercise:
Problem:
Exercise:
Problem:
Solution:
a=-4
Exercise:
Problem:
ae |
9
az _ 9
A "39
BY 85
3
3a 3 4
8 2
OT is 09 a
6 3
Exercises for Review
Exercise:
Problem: ([link]) Use the distributive property to compute 40 - 28.
Solution:
40(30 — 2) = 1200 — 80 = 1120
Exercise:
Problem:
({link]) Approximating 7 by 3.14, find the approximate circumference
of the circle.
( ee
Exercise:
Problem: ({link]) Find the area of the parallelogram.
20 cm
Solution:
220 sq cm
Exercise:
—3(4-—15)-—2
Problem: ({link]) Find the value of SS
Exercise:
Problem: ([link]) Solve the equation x — 14+ 8 = —2.
Solution:
rad
Applications I: Translating Words to Mathematical Symbols
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module discusses how to translate word to
mathematical symbols. By the end of the module students should be able to
translate phrases and statements to mathematical expressions and equations.
Section Overview
e Translating Words to Symbols
Translating Words to Symbols
Practical problems seldom, if ever, come in equation form. The job of the
problem solver is to translate the problem from phrases and statements into
mathematical expressions and equations, and then to solve the equations.
As problem solvers, our job is made simpler if we are able to translate
verbal phrases to mathematical expressions and if we follow the five-step
method of solving applied problems. To help us translate from words to
symbols, we can use the following Mathematics Dictionary.
MATHEMATICS DICTIONARY
Mathematical
Word or Phrase :
Operation
Sum, sum of, added to, increased by, more r
than, and, plus
Difference, minus, subtracted from, decreased
by, less, less than
Product, the product of, of, multiplied by,
times, per
Quotient, divided by, ratio, per =
Equals, is equal to, is, the result is, becomes =
A number, an unknown quantity, an unknown, x (or any
a quantity symbol)
Sample Set A
Translate each phrase or sentence into a mathematical expression or
equation.
Example:
Ninemore thansome number.
9 + xr
Translation: 9 + 2.
Example:
EKighteenminusa number.
18 = €
Translation: 18 — z.
Example:
A quantity less five.
—- 65
y
Translation: y — 5.
Example:
Fourtimesa number is sixteen.
4 : a8 = 16
Translation: 4x = 16.
Example:
One fifth of anumber is thirty.
1 . =
+ : 30
: 1 n
Translation: at =-S0 sor 5 =i 0b
Example:
Five times anumber is two more than twice the number.
5 ; ar ee = 2. ar
Translation: 52 = 2 + 22.
Practice Set A
Translate each phrase or sentence into a mathematical expression or
equation.
Exercise:
Problem: Twelve more than a number.
Solution:
12+ 2
Exercise:
Problem: Eight minus a number.
Solution:
8-2
Exercise:
Problem: An unknown quantity less fourteen.
Solution:
z—14
Exercise:
Problem: Six times a number is fifty-four.
Solution:
67-54
Exercise:
Problem: Two ninths of a number is eleven.
Solution:
2
—xz-—11
9
Exercise:
Problem:
Three more than seven times a number is nine more than five times the
number.
Solution:
3+ 72 =—9+ 52
Exercise:
Problem:
Twice a number less eight is equal to one more than three times the
number.
Solution:
2x —8 = 3z+1lor2x —-8 = 1+ 32
Sample Set B
Example:
Sometimes the structure of the sentence indicates the use of grouping
symbols. We’ll be alert for commas. They set off terms.
A numberdivided byfour,minus six, is twelve
6 = iy
(x + 4)
Translation: - —6= 12.
Example:
Some phrases and sentences do not translate directly. We must be careful to
read them properly. The word from often appears in such phrases and
sentences. The word from means “a point of departure for motion.” The
following translation will illustrate this use.
Twenty is subtracted from some number.
Neen omen? Vennseenmnsen yess Vettearpsnta set?
x 20
Translation: x — 20.
The word from indicated the motion (subtraction) is to begin at the point of
“some number.”
Example:
Ten less than some number. Notice that less than can be replaced by from.
Ten from some number.
Translation: x — 10.
Practice Set B
Translate each phrase or sentence into a mathematical expression or
equation.
Exercise:
Problem: A number divided by eight, plus seven, is fifty.
Solution:
4 0
~+7=50
8
Exercise:
Problem:
A number divided by three, minus the same number multiplied by six,
is one more than the number.
Solution:
2
-~ _ 67 = 1
3 LE-=at+
Exercise:
Problem: Nine from some number is four.
Solution:
r—-9-—4
Exercise:
Problem: Five less than some quantity is eight.
Solution:
zr—-5=—8
Exercises
Translate each phrase or sentence to a mathematical expression or equation.
Exercise:
Problem: A quantity less twelve.
Solution:
xz—12
Exercise:
Problem: Six more than an unknown number.
Exercise:
Problem: A number minus four.
Solution:
r—A4
Exercise:
Problem: A number plus seven.
Exercise:
Problem: A number increased by one.
Solution:
x+1
Exercise:
Problem: A number decreased by ten.
Exercise:
Problem: Negative seven added to some number.
Solution:
—7T+2
Exercise:
Problem: Negative nine added to a number.
Exercise:
Problem: A number plus the opposite of six.
Solution:
x + (—6)
Exercise:
Problem: A number minus the opposite of five.
Exercise:
Problem: A number minus the opposite of negative one.
Solution:
«—[—(-1)
Exercise:
Problem: A number minus the opposite of negative twelve.
Exercise:
Problem: Eleven added to three times a number.
Solution:
32+ 11
Exercise:
Problem: Six plus five times an unknown number.
Exercise:
Problem: Twice a number minus seven equals four.
Solution:
2x —7=—4
Exercise:
Problem: Ten times a quantity increased by two is nine.
Exercise:
Problem:
When fourteen is added to two times a number the result is six.
Solution:
14+ 27% = 6
Exercise:
Problem: Four times a number minus twenty-nine is eleven.
Exercise:
Problem: Three fifths of a number plus eight is fifty.
Solution:
3
= 8 = 50
5 4
Exercise:
Problem: Two ninths of a number plus one fifth is forty-one.
Exercise:
Problem:
When four thirds of a number is increased by twelve, the result is five.
Solution:
= +12=5
3
Exercise:
Problem:
When seven times a number is decreased by two times the number, the
result is negative one.
Exercise:
Problem:
When eight times a number is increased by five, the result is equal to
the original number plus twenty-six.
Solution:
8x1 +5=—2+4+ 26
Exercise:
Problem:
Five more than some number is three more than four times the number.
Exercise:
Problem:
When a number divided by six is increased by nine, the result is one.
Solution:
Ab
—+9=1
6
Exercise:
Problem: A number is equal to itself minus three times itself.
Exercise:
Problem: A number divided by seven, plus two, is seventeen.
Solution:
x
tae? cae Wh
71
Exercise:
Problem:
A number divided by nine, minus five times the number, is equal to
one more than the number.
Exercise:
Problem: When two is subtracted from some number, the result is ten.
Solution:
z—2=—10
Exercise:
Problem:
When four is subtracted from some number, the result is thirty-one.
Exercise:
Problem:
Three less than some number is equal to twice the number minus six.
Solution:
G—=—3=2e:—6
Exercise:
Problem:
Thirteen less than some number is equal to three times the number
added to eight.
Exercise:
Problem:
When twelve is subtracted from five times some number, the result is
two less than the original number.
Solution:
De 12 SS
Exercise:
Problem:
When one is subtracted from three times a number, the result is eight
less than six times the original number.
Exercise:
Problem:
When a number is subtracted from six, the result is four more than the
original number.
Solution:
6-—x=24+4
Exercise:
Problem:
When a number is subtracted from twenty-four, the result is six less
than twice the number.
Exercise:
Problem:
A number is subtracted from nine. This result is then increased by one.
The result is eight more than three times the number.
Solution:
J=—2-- 136-6
Exercise:
Problem:
Five times a number is increased by two. This result is then decreased
by three times the number. The result is three more than three times the
number.
Exercise:
Problem:
Twice a number is decreased by seven. This result is decreased by four
times the number. The result is negative the original number, minus
Six,
Solution:
2x —7—4x2 = -—x2 —-6
Exercise:
Problem:
Fifteen times a number is decreased by fifteen. This result is then
increased by two times the number. The result is negative five times
the original number minus the opposite of ten.
Exercises for Review
Exercise:
8 2
Problem: ({link}) a of what number is
Solution:
3
A
Exercise:
21. 1%
Problem: ({link]) Find the value of —- + —.
40 30
Exercise:
1 1 1
Problem: ({link]) Find the value of 375 + aa + 17.
Solution:
2
—
3
Exercise:
1
Problem: ({link]) Convert 6.11 5 to a fraction.
Exercise:
3
Problem: ({link]) Solve the equation = +1=-—-5.
Solution:
£==-8
Applications II: Solving Problems
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses
how to solve algebraic problems. By the end of the module students should be more familiar with the five-step
method for solving applied problems and be able to use the five-step method to solve number problems and
geometry problems.
Section Overview
e The Five-Step Method
¢ Number Problems
e¢ Geometry Problems
The Five Step Method
We are now in a position to solve some applied problems using algebraic methods. The problems we shall solve
are intended as logic developers. Although they may not seem to reflect real situations, they do serve as a basis for
solving more complex, real situation, applied problems. To solve problems algebraically, we will use the five-step
method.
Strategy for Reading Word Problems
When solving mathematical word problems, you may wish to apply the following "reading strategy." Read the
problem quickly to get a feel for the situation. Do not pay close attention to details. At the first reading, too much
attention to details may be overwhelming and lead to confusion and discouragement. After the first, brief reading,
read the problem carefully in phrases. Reading phrases introduces information more slowly and allows us to
absorb and put together important information. We can look for the unknown quantity by reading one phrase at a
time.
Five-Step Method for Solving Word Problems
1. Let x (or some other letter) represent the unknown quantity.
2. Translate the words to mathematical symbols and form an equation. Draw a picture if possible.
3. Solve the equation.
4. Check the solution by substituting the result into the original statement, not equation, of the problem.
5. Write a conclusion.
If it has been your experience that word problems are difficult, then follow the five-step method carefully. Most
people have trouble with word problems for two reasons:
1. They are not able to translate the words to mathematical symbols. (See [link].)
2. They neglect step 1. After working through the problem phrase by phrase, to become familiar with the
situation,
INTRODUCE A VARIABLE
Number Problems
Sample Set A
Example:
What number decreased by six is five?
Let”represent the unknown number.
Translate the words to mathematical symbols and construct an equation. Read
phrases.
What number: n
decreased by: —
Six: O M=G=F
is: =
five: 5
Solve this equation.
n — 6 = 5 Add 6 to both sides.
= C+ O=546
i = il
Check the result.
When 11 is decreased by 6, the result is 11 — 6, which is equal to 5. The solution checks.
The number is 11.
Example:
When three times a number is increased by four, the result is eight more than five times the number.
Let® =the unknown number.
Translate the phrases to mathematical symbols and
construct an equation. When three times a number: 32
is increased by: +
four: 4
the result is: 382 +4=524+8
eight: 8
more than: +
five times the number: 5a
32 +4=52r+ 8. Subtract3xfrom both sides.
3x2 +4-—32 =5274+8-—32
4=2x+8 Subtract 8 from both sides.
4—8=2x+8-8
—4 = 2x Divide both sides by 2.
= =
CheckThree—2is—6, =py4 —64+4— —2. 2is—10 Increasing—10by8results—10 + 8 = —2, TE
this times Increasing __ results Now, in rest
result. in five agre
times and
solu
che:
The number is—2
Example:
Consecutive integers have the property that if
n = the smallest integer, then
n+1 = the next integer, and
n+2 = the next integer, and so on.
Consecutive odd or even integers have the property that if
n = the smallest integer, then
n+2 = the next odd or even integer (since odd or even numbers differ by 2), and
n+4 = the next odd or even integer, and so on.
The sum of three consecutive odd integers is equal to one less than twice the first odd integer. Find the three
integers.
Let n = the first odd integer. Then,
n+2 = the second odd integer, and
n+4 = the third odd integer.
Translate the words
to mathematical
symbols and
construct an
equation. Read
phrases.
The sum of:
three consecutive odd integers:
is equal to:
one less than:
twice the first odd integer:
add some numbers
nn+2n+4
= a+ (n+ 2)4
subtract 1 from
2n
ntn+2+n+4=2n-1
3n+6=2n-1 Subtract2nfrom both sides.
3n +6 — 2n = 2n—1-—2n
n+6=-1 Subtract 6 from both sides.
n+6—-6=-1-6
ee — —T The first integer is -7.
DARA = =f ba = =F The second integer is -5.
(ae al ff th a 8} The third integer is -3.
Check The sum of
this the three —12 + (—3)One less than twice the first integer is
=f (0) (3)
2(—7) —1 = —14-—1 = —15. Since these two
results are equal, the solution checks.
result. integers is —15
The three odd integers are -7, -5, -3.
Practice Set A
Exercise:
Problem: When three times a number is decreased by 5, the result is -23. Find the number.
Let? =
Check:
The number is.
Solution:
-6
Exercise:
Problem:
When five times a number is increased by 7, the result is five less than seven times the number. Find the
number.
Let”? =
Check:
The number is.
Solution:
6
Exercise:
Problem: Two consecutive numbers add to 35. Find the numbers.
Check:
The numbers areand.
Solution:
17 and 18
Exercise:
Problem:
The sum of three consecutive even integers is six more than four times the middle integer. Find the integers.
Let® smallest integer.= next integer.= largest integer.
Check:
The integers are,, and.
Solution:
-8, -6, and -4
Geometry Problems
Sample Set B
Example:
The perimeter (length around) of a rectangle is 20 meters. If the length is 4 meters longer than the width, find the
length and width of the rectangle.
Let® =the width of the rectangle. Then,® + 4 =the length of the rectangle.
We can draw a picture.
x+4 The length around the rectangle is
e +(e¢+4)4+ 2 +(¢+4)=20
# x width length width length
x+4
Ar +8 = 20 Subtract 8 from both sides.
4p == 1 Divide both sides by 4.
B=3 Then,
r+4=3+4=7
Check:
7 38+7+3+7220
20 ~ 20
3 3
7
The length of the rectangle is 7 meters.The width of the rectangle is 3 meters.
Practice Set B
Exercise:
Problem:
The perimeter of a triangle is 16 inches. The second leg is 2 inches longer than the first leg, and the third leg
is 5 inches longer than the first leg. Find the length of each leg.
Let® length of the first leg.= length of the second leg.= length of the third leg.
We can draw a picture.
Check:
The lengths of the legs are,, and.
Solution:
3 inches, 5 inches, and 8 inches
Exercises
For the following 17 problems, find each solution using the five-step method.
Exercise:
Problem: What number decreased by nine is fifteen?
Let”? =the number.
Check:
The number is.
Solution:
24
Exercise:
Problem: What number increased by twelve is twenty?
n=
Let the number.
Check:
The number is.
Exercise:
Problem: If five more than three times a number is thirty-two, what is the number?
Let® =the number.
Check:
The number is.
Solution:
9
Exercise:
Problem: If four times a number is increased by fifteen, the result is five. What is the number?
Lett =
Check:
The number is.
Exercise:
Problem:
When three times a quantity is decreased by five times the quantity, the result is negative twenty. What is the
quantity?
Let® =
Check:
The quantity is.
Solution:
10
Exercise:
Problem:
If four times a quantity is decreased by nine times the quantity, the result is ten. What is the quantity?
Let¥ =
Check:
The quantity is.
Exercise:
Problem:
When five is added to three times some number, the result is equal to five times the number decreased by
seven. What is the number?
Let”? =
Check:
The number is.
Solution:
6
Exercise:
Problem:
When six times a quantity is decreased by two, the result is six more than seven times the quantity. What is
the quantity?
Let? =
Check:
The quantity is.
Exercise:
Problem:
When four is decreased by three times some number, the result is equal to one less than twice the number.
What is the number?
Check:
Solution:
1
Exercise:
Problem:
When twice a number is subtracted from one, the result is equal to twenty-one more than the number. What is
the number?
Exercise:
Problem:
The perimeter of a rectangle is 36 inches. If the length of the rectangle is 6 inches more than the width, find
the length and width of the rectangle.
Let’ =the width.= the length.
We can draw a picture.
Check:
The length of the rectangle isinches, and the width isinches.
Solution:
Length=12 inches, Width=6 inches
Exercise:
Problem:
The perimeter of a rectangle is 48 feet. Find the length and the width of the rectangle if the length is 8 feet
more than the width.
Let’ =the width.= the length.
We can draw a picture.
Check:
The length of the rectangle isfeet, and the width isfeet.
Exercise:
Problem: The sum of three consecutive integers is 48. What are they?
Let” =the smallest integer.= the next integer.= the next integer.
Check:
The three integers are,, and.
Solution:
15, 16, 17
Exercise:
Problem: The sum of three consecutive integers is -27. What are they?
Let” =the smallest integer.= the next integer.= the next integer.
Check:
The three integers are,, and.
Exercise:
Problem: The sum of five consecutive integers is zero. What are they?
Let”? =
The five integers are,,,, and.
Solution:
-2,-1,0, 1,2
Exercise:
Problem: The sum of five consecutive integers is -5. What are they?
Let? =
The five integers are,,,, and.
Continue using the five-step procedure to find the solutions.
Exercise:
Problem:
The perimeter of a rectangle is 18 meters. Find the length and width of the rectangle if the length is 1 meter
more than three times the width.
Solution:
Length is 7, width is 2
Exercise:
Problem:
The perimeter of a rectangle is 80 centimeters. Find the length and width of the rectangle if the length is 2
meters less than five times the width.
Exercise:
Problem:
Find the length and width of a rectangle with perimeter 74 inches, if the width of the rectangle is 8 inches less
than twice the length.
Solution:
Length is 15, width is 22
Exercise:
Problem:
Find the length and width of a rectangle with perimeter 18 feet, if the width of the rectangle is 7 feet less than
three times the length.
Exercise:
Problem:
A person makes a mistake when copying information regarding a particular rectangle. The copied information
is as follows: The length of a rectangle is 5 inches less than two times the width. The perimeter of the
rectangle is 2 inches. What is the mistake?
Solution:
The perimeter is 20 inches. Other answers are possible. For example, perimeters such as 26, 32 are possible.
Exercise:
Problem:
A person makes a mistake when copying information regarding a particular triangle. The copied information
is as follows: Two sides of a triangle are the same length. The third side is 10 feet less than three times the
length of one of the other sides. The perimeter of the triangle is 5 feet. What is the mistake?
Exercise:
Problem:
The perimeter of a triangle is 75 meters. If each of two legs is exactly twice the length of the shortest leg, how
long is the shortest leg?
Solution:
15 meters
Exercise:
Problem:
If five is subtracted from four times some number the result is negative twenty-nine. What is the number?
Exercise:
Problem: If two is subtracted from ten times some number, the result is negative two. What is the number?
Solution:
n=0
Exercise:
Problem:
If three less than six times a number is equal to five times the number minus three, what is the number?
Exercise:
Problem:
If one is added to negative four times a number the result is equal to eight less than five times the number.
What is the number?
Solution:
ab
Exercise:
Problem: Find three consecutive integers that add to -57.
Exercise:
Problem: Find four consecutive integers that add to negative two.
Solution:
-2,-1,0,1
Exercise:
Problem: Find three consecutive even integers that add to -24.
Exercise:
Problem: Find three consecutive odd integers that add to -99.
Solution:
-35, -33, -31
Exercise:
Problem:
Suppose someone wants to find three consecutive odd integers that add to 120. Why will that person not be
able to do it?
Exercise:
Problem:
Suppose someone wants to find two consecutive even integers that add to 139. Why will that person not be
able to do it?
Solution:
...because the sum of any even number (in this case, 2) o even integers (consecutive or not) is even and,
therefore, cannot be odd (in this case, 139)
Exercise:
Problem:
Three numbers add to 35. The second number is five less than twice the smallest. The third number is exactly
twice the smallest. Find the numbers.
Exercise:
Problem:
Three numbers add to 37. The second number is one less than eight times the smallest. The third number is
two less than eleven times the smallest. Find the numbers.
Solution:
2, 15, 20
Exercises for Review
Exercise:
Problem: ((link]) Find the decimal representation of 0.34992 + 4.32.
Exercise:
Problem:
([link]) A 5-foot woman casts a 9-foot shadow at a particular time of the day. How tall is a person that casts a
10.8-foot shadow at the same time of the day?
Solution:
6 feet tall
Exercise:
5 1
Problem: ({link]) Use the method of rounding to estimate the sum: oe + be.
Exercise:
Problem: ({link]) Convert 463 mg to cg.
Solution:
46.3 cg
Exercise:
Problem:
((link]) Twice a number is added to 5. The result is 2 less than three times the number. What is the number?
Summary of Key Concepts
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module reviews the key concepts from the chapter
"Algebraic Expressions and Equations."
Summary of Key Concepts
Numerical Expression ({link])
A numerical expression results when numbers are associated by arithmetic
operation signs. The expressions 3 + 5, 9 — 2,5-6 and 8 + 5are
numerical expressions.
Algebraic Expressions ((link])
When an arithmetic operation sign connects a letter with a number or a
letter with a letter, an algebraic expression results. The expressions 4x7 + 1
,x—9, 7x - by, and 4x + 3 are algebraic expressions.
Terms and Factors ({link])
Terms are parts of sums and are therefore separated by addition (or
subtraction) signs. In the expression, 5x — 2y, 5x and —2y are the terms.
Factors are parts of products and are therefore separated by multiplication
signs. In the expression 5a, 5 and a are the factors.
Coefficients ({link])
The coefficient of a quantity records how many of that quantity there are.
In the expression 7z, the coefficient 7 indicates that there are seven z's.
Numerical Evaluation ([link])
Numerical evaluation is the process of determining the value of an
algebraic expression by replacing the variables in the expression with
specified values.
Combining Like Terms ({link])
An algebraic expression may be simplified by combining like terms. To
combine like terms, we simply add or subtract their coefficients then affix
the variable. For example 4x + 92 = (4+ 9)ax = 132.
Equation ((link])
An equation is a statement that two expressions are equal. The statements
4 2
52+ 1= 3 and = +4= 5 are equations. The expressions represent the
Same quantities.
Conditional Equation ((link])
A conditional equation is an equation whose truth depends on the value
selected for the variable. The equation 32 = 9 is a conditional equation
since it is only true on the condition that 3 is selected for z.
Solutions and Solving an Equation ((link])
The values that when substituted for the variables make the equation true
are called the solutions of the equation.
An equation has been solved when all its solutions have been found.
Equivalent Equations ({link])
Equations that have precisely the same solutions are called equivalent
equations. The equations 6y = 18 and y = 3 are equivalent equations.
Addition/Subtraction Property of Equality ({link])
Given any equation, we can obtain an equivalent equation by
1. adding the same number to both sides, or
2. subtracting the same number from both sides.
Solving x + a = band z — a = 6 ([link))
To solve xz + a = Db, subtract a from both sides.
rta=—b
xrt+a-a=b-a
xz—b-a
To solve x — a = b, add a to both sides.
xrz—a=b
zr—-at+a=—ba
z—b+a
Multiplication/Division Property of Equality ({link])
Given any equation, we can obtain an equivalent equation by
1. multiplying both sides by the same nonzero number, that is, if c 4 0,
a = banda-c= 0 - care equivalent.
2. dividing both sides by the same nonzero number, that is, if c ~ 0,
a :
a = band — = — are equivalent.
‘a c
Solving az = b and a b ({link])
a
To solve ax = b, a £ 0, divide both sides by a.
in = 4%
ax ti(‘i‘“C
a — @
Ae ob
v4 ~— @
c= +
a
To solve a b, a ~ 0, multiply both sides by a.
a
- = 6b
a-*~ = a-b
aad = Gh
ao = jae
Translating Words to Mathematics ([link])
In solving applied problems, it is important to be able to translate phrases
and sentences to mathematical expressions and equations.
The Five-Step Method for Solving Applied Problems ((link])
To solve problems algebraically, it is a good idea to use the following five-
step procedure.
After working your way through the problem carefully, phrase by phrase:
1. Let x (or some other letter) represent the unknown quantity.
2. Translate the phrases and sentences to mathematical symbols and form
an equation. Draw a picture if possible.
3. Solve this equation.
4. Check the solution by substituting the result into the original statement
of the problem.
5. Write a conclusion.
Exercise Supplement
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is an exercise supplement for the chapter
"Algebraic Expressions and Equations" and contains many exercise
problems. Odd problems are accompanied by solutions.
Exercise Supplement
Algebraic Expressions ((link])
For problems 1-10, specify each term.
Exercise:
Problem: 6a — 2b + 5c
Solution:
6a, —25, 5c
Exercise:
Problem: 9x — 6y + 1
Exercise:
Problem: 7m — 3n
Solution:
7™m, —3n
Exercise:
Problem: —5h + 2k —8+4m
Exercise:
Problem:
Solution:
x, 2n, —z
Exercise:
Problem:
Exercise:
Problem:
Solution:
Y= Bz
Exercise:
Problem:
Exercise:
Problem:
Solution:
—4
Exercise:
Problem:
Exercise:
Problem:
r+2n-—z
y—9
—y — 3z
=—a—b—¢c-—1
—4
—6
Write 1k in a simpler way.
Solution:
k
Exercise:
Problem: Write 1z in a simpler way.
Exercise:
Problem: In the expression 7r, how many r’s are indicated?
Solution:
7
Exercise:
Problem: In the expression 12™m, how many m’s are indicated?
Exercise:
Problem: In the expression —5n, how many n’s are indicated?
Solution:
-5
Exercise:
Problem: In the expression —10y, how many y’s are indicated?
For problems 17-46, find the value of each expression.
Exercise:
Problem: 5a — 2s, ifa = —5bands=1
Solution:
-27
Exercise:
Problem:
Exercise:
Problem:
Solution:
1
Exercise:
Problem:
Exercise:
Problem:
Solution:
-5
Exercise:
Problem:
Exercise:
Problem:
Solution:
0
Exercise:
Tn — 3r, ifn = —6 andr = 2
9x + 2y — 3s, ifx = —2,y = 5, ands = —3
10a — 2b-+ 5c, ifa — 0, b = —6, andc = 8
—5s — 2t+1,ifs = 2 andt = —2
—3m —4n+5,ifm = —landn = —1
m—4,ifm=4
Problem
Exercise:
Problem:
= 2. tS
=f + 2y, 1 e=>—Candy=—1
Solution:
s)
Exercise:
Problem:
Exercise:
Problem:
—a+3b—6,ifa = —3andb=0
5a —4y —Ty+y— 72, ife = 1 andy = —2
Solution:
18
Exercise:
Problem:
Exercise:
Problem:
2a — 6b — 3a—a-+ 2b, ifa = 4andb = —2
a” —6a+4,ifa = —2
Solution:
20
Exercise:
Problem:
m? — 8m — 6, ifm = —5
Exercise:
Problem:
Solution:
11
Exercise:
Problem:
Exercise:
Problem:
Solution:
“if
Exercise:
Problem:
Exercise:
Problem:
Solution:
-93
Exercise:
Problem:
Exercise:
Ay”? + 3y +1, if y = —2
5a’ — 6a +11, ifa —0
—k*?-—k-1,ifk=-1
She Ih= 3 ith=]—4
= +5m, ifm = —18
5 2a+lifa =24
Problem:
Solution:
45
Exercise:
Problem:
Exercise:
Problem:
Solution:
eye)
Exercise:
Problem:
Exercise:
Problem:
Solution:
10
Exercise:
Problem:
5
+32 —-Tife =14
3k
7 —5k+ 18, if k = 16
_6
—* + 3a +10, ifa = 25
—* —th-1,ifh = -18
5(3a + 4b), ifa = —2 andb = 2
7(2y — x), if x = —landy = 2
Exercise:
Problem: —(a — b), if a= 0 andb = —6
Solution:
-6
Exercise:
Problem: —(x — x — y), ifz = 4 andy = —4
Exercise:
Problem: (y + 2)? — 6(y + 2) — 6, ify = 2
Solution:
-14
Exercise:
Problem: (a — 7)? — 2(a — 7) — 2, ifa =7
Combining Like Terms Using Addition and Subtraction ((link])
For problems 47-56, simplify each expression by combining like terms.
Exercise:
Problem: 4a + 5 — 2a+ 1
Solution:
2a+6
Exercise:
Problem: 7x + 3x — 14x
Exercise:
Problem: —7b + 4m — 3+ 3n
Solution:
—4n+4m— 3
Exercise:
Problem: —9k — 8h —k+ 6h
Exercise:
Problem: —z + 5y — 8x — 6x + Ty
Solution:
—15z + 12y
Exercise:
Problem: 6n — 2n+6—-2-n
Exercise:
Problem: 0m + 3k — 5s + 2m—s
Solution:
3k + 2m — 6s
Exercise:
Problem: | —8 | a+ | 2 | b— | —4|a
Exercise:
Problem: | 6 | h— | —7|k+|—12|h+|4|-|—5|h
Solution:
38h — 7k
Exercise:
Problem: | 0 | a— 0a+0
ep
Equations of the Form az = b and — = 6), Translating Words to
a
Mathematical Symbols , and Solving Problems ((link],[link],[link])
For problems 57-140, solve each equation.
Exercise:
Problem: z + 1=5
Solution:
pHa
Exercise:
Problem: y — 3 = —7
Exercise:
Problem: x + 12 = 10
Solution:
r= -—2
Exercise:
Problem:
Exercise:
Problem:
Solution:
p= 5
Exercise:
Problem:
Exercise:
Problem:
Solution:
pal
Exercise:
Problem:
Exercise:
Problem:
Solution:
z—-—l15
Exercise:
x—-4=—6
oe: 25
a2 = 17
ae
2
aoe
—8
OG
15
Problem:
Exercise:
Problem:
Solution:
C= 3
Exercise:
Problem:
Exercise:
Problem:
Solution:
‘aD
Exercise:
Problem:
Exercise:
Problem:
Solution:
x= —27
Exercise:
—32 —9
—22
—5ar
—3r2 — -l
Problem: = = 2
Exercise:
Problem: —7 = 3y
Solution:
Pe
Exercise:
xr
Problem: —7 = 5
Exercise:
—2
Problem: ids = —
A 5
Solution:
m=- =
5
Exercise:
1
Problem: 4y = 2
Exercise:
—l
Problem: — = —)5z
Solution:
2
15
Exercise:
x
Problem:
Exercise:
Problem:
Solution:
s—l
Exercise:
Problem:
Exercise:
Problem
Solution:
zr—-—3
Exercise:
Problem:
Exercise:
Problem:
<1 ik
9 8
= Ss
6 6
0
— = As
A
:2+2=—-l1
x—-5=—6
—3
—2z-—6
Solution:
x= —4
Exercise:
Problem: 3x + 2 = 7
Exercise:
Problem: —4z — 5 = —3
Solution:
La
Exercise:
Problem:
Exercise:
Problem: —— — 3 = —2
Solution:
C=—5
Exercise:
Problem: — = 7
Exercise:
Problem:
Solution:
joanne (25
Exercise:
Problem:
Exercise:
Problem:
Solution:
H ee 5
Exercise:
Problem:
Exercise:
Problem:
Solution:
es
=D
Exercise:
a
—+2=8
5 +
Lae eae
2
m+3=8
1x
Paes)
2
2
2a _
2
Problem:
Exercise:
Problem:
Solution:
=e
Exercise:
Problem:
Exercise:
Problem:
Solution:
i
Exercise:
Problem:
Exercise:
Problem:
Solution:
eae
Ses et
rf
ae eT,
—2
—4k-6=7
—32x
SS aa
=F 25
4
r+9=14
Exercise:
Problem: y + 5 = 21
Exercise:
Problem: y + 5 = —7
Solution:
y= —12
Exercise:
Problem: 4x2 = 24
Exercise:
Problem: 4w = 37
Solution:
3T
- 4
Exercise:
W
Problem: 6y — 11 = 13
Exercise:
Problem: —3z + 8 = —7
Solution:
r= 5
Exercise:
Problem: 3z + 9 = —5l
Exercise:
Problem: si —8
—3
Solution:
x2 — —24
Exercise:
6
Problem: ae. =
7
Exercise:
Problem: 5 —15=—4
Solution:
w — 38
Exercise:
“
Problem: =a 23 = —10
Exercise:
2
Problem: = —5-—8
Solution:
Exercise:
32 =7
Problem: — = ——
A 8
Exercise:
2
Problem: —2 — > =3
Solution:
35
C=
2
Exercise:
Problem: 3 — x = 4
Exercise:
Problem: —5 — y = —2
Solution:
ps8
Exercise:
Problem: 3 — z = —2
Exercise:
Problem: 3x + 2x = 6
Solution:
Se
Exercise:
Problem: 4x + 1+ 6z = 10
Exercise:
Problem: 6y — 6 = —4+ 3y
Solution:
y~ 3
Exercise:
Problem: 3 = 4a — 2a+a
Exercise:
Problem: 3m + 4= 2m+1
Solution:
m==3
Exercise:
Problem: 5w — 6 = 4+ 2w
Exercise:
Problem: 8 — 3a = 32 — 2a
Solution:
a— —24
Exercise:
Problem: 5z — 2x + 62 = 13
Exercise:
Problem: «+ 2= 3-2
Solution:
ea
Exercise:
Problem: 5y + 2y — 1 = 6y
Exercise:
Problem: x = 32
Solution:
p32
Exercise:
Problem: k = —4
Exercise:
5 5
Problem: bck +4= shee 6
2 2
Solution:
= 2
Exercise:
Problem:
Exercise:
Problem:
Solution:
ged
Exercise:
Problem:
Exercise:
Problem:
Solution:
S65
Exercise:
Problem:
Exercise:
Problem:
Solution:
L 32
ee ee ae
a 3
r—-2-—6- 2
—5a 2a
=) ae
BE desi
; =
—3x 22
2 Ae ALD
5 = 5 =
32 32
ES ge el FA |
4 = 4
c= a
Exercise:
3 —3
Problem: cd = = +12
4 7
Exercise:
SY Ty
Problem: — — 4 = — +1
roblem 3 26 ie
Solution:
_ 130
no ae
Exercise:
—3 6
Problem: Basi = Bee Es 2
5 10
Exercise:
—3
Problem: > +1—=—5m
Solution:
2
m= —
13
Exercise:
22
Problem: —3z = =
Proficiency Exam
This module is from Fundamentals of Mathematics by Denny Burzynski
and Wade Ellis, Jr. This module is a proficiency exam to the chapter
"Algebraic Expressions and Equations." Each problem is accompanied with
a reference link pointing back to the module that discusses the type of
problem demonstrated in the question. The problems in this exam are
accompanied by solutions.
Proficiency Exam
For problems 1 and 2 specify each term.
Exercise:
Problem: (({link]) 5a + 6y + 3z
Solution:
5x, by, 3z
Exercise:
Problem: ([link]) 8m — 2n — 4
Solution:
8m, —2n, -4
Exercise:
Problem: ({link]) In the expression —9a, how many a’s are indicated?
Solution:
-9
For problems 4-9, find the value of each expression.
Exercise:
Problem: ([link]) 6a — 3b, if a = —2, and b = —1.
Solution:
-9
Exercise:
Problem: ([link]) —5m + 2n — 6, ifm = —landn = 4.
Solution:
7
Exercise:
Problem: ((link]) —2? + 3x2 — 5, if = —2.
Solution:
-15
Exercise:
Problem: ((link]) y* + 9y + 1, if y = 0.
Solution:
1
Exercise:
Problem: ({link]) —a? + 3a + 4, ifa = 4.
Solution:
0)
Exercise:
Problem:
(Hlink]) —(5 — 2)*+7(m—2)+2—2m,ife =5andm=5.
Solution:
5
For problems 10-12, simplify each expression by combining like terms.
Exercise:
Problem: ((link]) 6y + 5 — 2y+ 1
Solution:
Ay +6
Exercise:
Problem: ((link]) 14a — 3b + 5b — 6a — b
Solution:
8a+b
Exercise:
Problem: ([link]) 9x + 5y — 7 + 4x” — 6y + 3(-2)
Solution:
32 —y—13
For problems 13-22, solve each equation.
Exercise:
Problem: ({link]) x + 7 = 15
Solution:
2r=8
Exercise:
Problem: ((link]) y — 6 = 2
Solution:
y=s
Exercise:
Problem: ({link]) m+ 8 = —1
Solution:
m==-9
Exercise:
Problem: ([link]) —5 +a = —4
Solution:
a=
Exercise:
Problem: ([link]) 4z = 104
Solution:
= 26
Exercise:
Problem
: ([link]) 6y + 3 = —21
Solution:
ve
Exercise:
Problem
Solution:
m— 4
Exercise:
Problem
Solution:
tea.
Exercise:
Problem
: (link) “= 2
onnieat Oe 2 le
nk) a rl
: ({link]) 62 +5 = 42 — 11
Solution:
£==8
Exercise:
Problem:
({link]) 4y — 8 — 6y = 8y+ 1
Solution:
—:
oe
Exercise:
Problem:
({link] and [link]) Three consecutive even integers add to -36. What
are they?
Solution:
-14, -12, -10
Exercise:
Problem:
({link] and [link]) The perimeter of a rectangle is 38 feet. Find the
length and width of the rectangle if the length is 5 feet less than three
times the width.
Solution:
l= 13,0 =6
Exercise:
Problem:
({link] and [link]) Four numbers add to -2. The second number is three
more than twice the negative of the first number. The third number is
six less than the first number. The fourth number is eleven less than
twice the first number. Find the numbers.
Solution:
6, -9, 0, 1
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